So nice of you dear! You are awesome 👍 I'm glad you liked it! Please keep sharing premath channel with your family and friends. Take care dear and stay blessed😃
I began college at 51 years of age and cried everyday in algebra. It was a foreign language to me. I'm 70 now and wish I would have knew about you then. I finally quit college after my third year because I felt so dumb 😰. I promised my mom I would graduate. She passed two years ago and I think I can accomplish college once I move out of state. Especially with your lessons.
We can cube both sides of our equation. (a-b)^3=a^3-b^3-3ab(a-b) a=cuberoot(x+28) b=cuberoot(x-28) a-b=2 (x+28)-(x-28)-3cuberoot[(x+28)(x-28)]*2=8 56-3(cuberoot(x^2-28^2)*2)=8 48-6(cuberoot(x^2-28^2))=0 8-cuberoot(x^2-28^2)=0 cuberoot(x^2-28^2)=8 x^2-28^2=8^3 x^2-784=512 x^2=1296 x=36 or x=-36
very cool problem that turned out a lot simpler than expected. here's another solution: ∛(x+28) - ∛(x-28) = 2 ∛(x+28) = 2 + ∛(x-28) x + 28 = (2 + ∛(x-28))^3 let u = ∛(x-28) the right hand side then becomes: (2 + u)^3 = (4 + 4u + u^2)(2+u) = 8 + 4u + 8u + 4u^2 + 2u^2 + u^3 = 8 + 12u + 6u^2 + u^3. now our equation is x+28 = 8 + 12u + 6u^2 + u^3 however, u^3 is just x - 28, so we can cancel x from both sides and get 6u^2 + 12u - 48 = 0 which can be factored into 6(u-2)(u+4) thus, u = 2 or u = -4 substituting u = ∛(x-28): 1: x - 28 = 2^3 ==> x = 36 2: x - 28 = (-4)^3 ==> x = -36 final answer: x = ± 36
I try to guess the solution since the RHS is just 2. may be the radicals are perfect cube roots 1³=1, 2³=8, 3³=27, 4³=64.... Then guess x+28=64 then x=36 substitute that back into the given equation verify x=36 is indeed correct. Since I know we are dealing with cube roots, negative numbers can be solution as well. ∵ (-1)³=-1, (-2)³=-8, (-3)³=-27, (-4)³=-64 ∴ I try to put x=-36 into the given equation and check it out and bingo "yes" so x can be ±36 x=±36
The solution should've been shorter: If a = cbrt(x+28), b = cbrt(x-28), and a - b = 2, then (a-b)^3 = a^3 - b^3 - 3ab(a-b) 8 = 56 - 6ab ab = 8 --> (ab)^3 = x^2 - 28^2 = 512 x = ±√(512 + 784) = ±36
Hello I was stunned by this problem because I didn’t think that substitution would help at all! In anyway which I would try to solve I would always end up with even more complex equation your way of solving was indeed very need and new! I solved 2nd degree radicals by moving one to the other side then squaring so here I moved to the other side and cubes (cause it’s degree 3) but in the end it would still have cube roots in it thank you for teaching me something new :D
Dear RS, this is a very challenging problem. When I was preparing this problem, I did it many different ways and finally decided to go with the method that I shared in the vid. Thanks dear fro sharing your experience. You are awesome 👍 Take care dear and stay blessed😃 Kind regards Love and prayers from Arizona, USA!
Great , step - by - step solution of a complex solution. It would be difficult to solve without writing out each step to keep track of all the information created by each step. It is also a test of knowledge of binomial expansions and manipulation of information created by each step. Mastery of this skill is a worthy goal. Thank you for the practice!
You have a-b=2 and ab=8 You can arrange a quadratic equation, for instance a^2-2a-8=0 whose solutions are a1=-2 and a2=4 And, of course; b1=-4 b2=2 Using a1=-2 will bring x=36 whereas a2=4 gives the solution x=-36
Another way would be to solve the equation graphically by y = (3rt x + 28) - (3rt x -28) + 2 and solve when y = 0. Plotting the equation I find with these complex equations is better to understand as it shows it visually. A good advantage of having a graphical calculator
Let y=cbrt(x+28)-1, then should be y-1=cbrt(x-28), 56=(x+28)-(x-28)=(y+1)^3-(y-1)^3=6y^2+2, 6y^2=54, y^2=9, y=3 or -3. If y=3, then cbrt(x-28)=2, x-28=8, x=36. If y=-3, then cbrt(x+28)=-2, x+28=-8, x=-36.
I applied the cubic formula directly resultantly left with the product of terms under radicals. I took the cube of the equation. My final equation was (x+28)(x-28)=512. Solving for x yielded x=36 and x=-36
Thanks Zaki Naqvi dear for the visit! Thanks for sharing You are awesome 👍 Take care dear and stay blessed😃 Kind regards Love and prayers from Arizona, USA!
Very intereseting solution. Apparently it can be simplified. In step 2 instead of calculating cube of difference, let’s work out square of difference (a-b) ^2=a^2+b^2-2ab=4. Subctract this equation from equation 3 and get ab=8, then multiply a and b: 3sqrt(x^2-784)=8, x=±36.
This is an interesting problem, but the solution has extraneous equations Once you have equations 1 and 3, you can substitute a=b+2 (obtained from equation 1) into the equation a^2+ab+b^2=28. The resulting simplified equation is b^2+2b-8=0, which is quite simple to solve for b.
Awesome question, nice solution! 😊 But here is a faster way: 2 x 28 = 56 = distance between (x + 28) and (x - 28). Because of „2“ on the right site, (x - 28) has to be 8 = 2³. 8 + 56 = 64 = 4³. Therefore: x = 36 (28 + 36 = 64; 36 - 28 = 8). Did it in 2 minutes 😊
1:46 it would make more sense if you crossed out the index number you know, instead of the top of the radical. That is nothing that needs to be crossed out. There is a lot of substitution going on here! This is one of my most favorite math problems out of all the other ones that you upload on TH-cam. I like Logarithms as well! Maybe, you could make up a logarithmic problem that is similar to this! That is my suggestion!
Again, I used a completely different method, but surprisingly got the right answers. I started by moving the second radical to the right side of the equation, then cubed both sides. Then a lot of algebra was required. The key was getting to a quadratic equation in which the variable being squared was a cube root! But from there it worked out very cleanly. I thought my method was cumbersome, but after seeing the official solution, I'm not so sure. I solved for x directly, with no substitution of variables used.
Great job J R. Thanks for sharing your experience. No matter what method you use, it requires extensive algebra. When I was preparing this problem, I thought about using your method as well. Anyway, thanks for your candid feedback. You are awesome 👍 Take care dear and stay blessed😃
I was thought the same way as the person below said about moving the cube root to the other side. This method is a little longer but love it because I have trouble with u substitution in calculus and need more practice. Thank you very much for this video always great to learn a different method, anther tool in the toolbox. 👍
Thanks Adeline for the visit! I put my heart and soul in this video. You are awesome 👍 Take care dear and stay blessed😃 Kind regards Love and prayers from Arizona, USA!
Sir, With ur substitution we ve, x= 1/2(a^3 + b^3) a-b=2 a^3 - b^3 =56 Using last two eqns.. We can have a quadratic equation in b. Solve this to get b and hence a. Use these values of a and b in first eqn to get x.
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Write down the first few cube numbers. Find two of them that have a difference of 56. There they are. 64 and 8. Or the corresponding negatives. Check that both pairs work. Done.
That is just guessing and what if the answer was some fraction of what if the answer was a cube or square root too! Then your method wouldn’t work so it’s best to solve systemically
Finding the difference between (x + 28) and (x - 28) is not a guess. It is a calculation. A good mathematician inspects a problem for a feature that might lead to a simple elegant solution. Of course my posting above would need to be expanded into a rigorous solution.
Finding value of a,b simply this way .we had a-b=2 and ab=8. So,a+b=under root (a-b)square+4ab which is under rt(4+32)=+-6 giving the value of a,b as shown.
Start with (x+28)^1/3=(x-28)^1/3+2 and take cube of both sides. After simplifications you are left with quadratic equation where (x-28)^1/3 is either 2 or -4.
We only need to consider positive x, because if a solution is found to the equation for +x, then -x is also a solution. As positive x increases, then ∛(x + 28) - ∛(x - 28) is decreasing. We notice at x = 0, then ∛(x + 28) - ∛(x - 28) = 2∛28 > 2∛27 = 2*3 = 6 and the difference is decreasing towards 0, but is always positive, except at infinite, where the difference is zero. So, if heuristically we assume that x is a positive integer, then we could try and find a positive cube, say a³, such that x + 28 = a³, where x > 0 to try and find the solution. The smallest a that meets the above conditions is a = 4 ⇒ a³ = 64 ⇒ x = 36 as a solution, and as -x is also a solution, then x = -36 is another solution. Have we found all the solutions? Yes, as ∛(x + 28) - ∛(x - 28) is decreasing and can only be equal to 2 "once", as positive x increases in value.
another way to solve this is just to cube both side or the original equation. (x+28) - (x - 28) - 3 cbrt(x+28) cbrt(x-28) (cbrt(x+28) - cbrt(x-28) ) = 8. which can be simplified as cbrt(x+28) cbrt(x-28) = 8, cube both side again x^2 - 28^2 = 8^3 , and will get x = +/- 36.
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When I looked at this problem I just didn't know where to start. By considering what the graph of the function looked like I concluded there would be 2 solutions: +X and -X, but more than that I was stuck. Putting the left hand size as a - b in hind sight was obvious but I didn't think of it - once you do it all falls out really nicely.
Set u=cube root of (x+28), v=cube root of (x-28). Therefore: u^3=x+28, v^3=x-28 So, we have a system of two equations u-v=2 and u^3-v^3=56. This system of two equation has two solutions, they are u=4, v=2 and u=-2, v=-4. With u=4, we have x=36, And with u=-2, we have x=-36. Summarising, we have two solutions, they are x=36 and x=-36.
I solved this a completely different way. Spotted The difference between a and b is 2 and the difference between a^3 and b^3 is 56. Logic said the only way to get these numbers were 2 ,4 and 8,64. Mine was logic - wouldn’t work if it wasn’t an integer problem though. Is my solution valid?
I did the same. LOL Sometimes it's best to take a shot at the easiest solution. Even if it turned out to not be an integer, you'd have a decent bound on the correct answer.
Looking at the problem, i know the solution is long. Those who dislikes math hates long solution. Others just don't know where to start. Math is practice and those who are good in math have their own method in solving math problems. Having said that there are more than one way to solve a math problem. Personally, my method is automatic the moment I see a math problem. In worded problems though, the first 4 steps are write the given, then what the problem is asking, then draw a diagram or figure and fourth the solution. The solution is always base on your analysis of the figure or diagram.
Let u = x - 28, so u + 56 = x + 28 Place these in the original equation and you get ∛(u + 56) - ∛u = 2 ⇒ ∛(u + 56) = 2 + ∛u ⇒ u + 56 = (2 + ∛u)³ = 8 + 3*2² * ∛u + 3*2¹ * (∛u)² + (∛u)³ = 8 + 12∛u + 6(∛u)² + u ⇒ 6(∛u)² + 12∛u - 48 = 0 Let y = ∛u and we end up with a quadratic equation 6y² + 12y - 48 = 0, which has only two solution for y, namely y = 2 or y = -4, which leads to u = 8 or u = -64. u = 8 and u = x -28 ⇒ x = 36 u = -64 and u = x - 28 ⇒ x = -36. So, the quadratic equation shows we have two solution. Now if the problem didn't have the same coefficients of x under the cube root signs, then there would be three solutions but perhaps not all real and not integers.
A very wierd function to work with... I let f(x)=(x+28)^(1/3) and g(x)=(x-28)^(1/3)... then carried out with the pattern of which f(x)-g(x)=2. I recirved 2 vertical lines; x= 36 and x=-36. 🤔
Ok, that was a bit harder than I thought. So I rearrange: sq3(x+28) = 2 + sq3(x-28) I raise both sides to the cube : x+28 = 8 + 3*4* sq3(x-28) + 3*2*sq3(x-28)² + (x-28) So 48 = 3*4* sq3(x-28) + 3*2*sq3(x-28)² 8 = 2* sq3(x-28) + sq3(x-28)² I pose y = sq3(x-28) and I get y²+2y-8 = 0 So y = 2 or y = -4 As x-28 is negative, y must be negative too, so y = -4 ( Edit: Wrong assumption, see answer below) so y == sq3(x-28) = -4 So I raise to the cube and I get: x-28 = -64 x = -36 Let's check The initial expression evaluates to sq3(-36+28) - sq3(-36-28) = sq3(-8) - sq3(-64) = 2 - (-4) = 2 Which confirms my solution.
Ah, it seems I missed the other solution which is x=36 I'm not sure where I lost that solution along the way. Oh yes, I can't just say that x-28 is negative... So I should test y = 2 Which give x-28 = 8 So x = 36 Somehow I find my solution easier to follow but that is probably just my brain being biased in favor of itself.
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So nice of you Jayant dear! You are awesome 👍 I'm glad you liked it! Please keep sharing premath channel with your family and friends. Take care dear and stay blessed😃
In H.S.L.C final exam that is class X (Assam, India) 1983 similar question was given with 2 marks. Unfortunately, I failed to solve. Thanks a lot for your maths classes.
Let u = x - 28, so u + 56 = x + 28 Place these in the original equation and you get ∛(u + 56) - ∛u = 2 ⇒ ∛(u + 56) = 2 + ∛u ⇒ u + 56 = (2 + ∛u)³ = 8 + 3*2² * ∛u + 3*2¹ * (∛u)² + (∛u)³ = 8 + 12∛u + 6(∛u)² + u ⇒ 6(∛u)² + 12∛u - 48 = 0 Let y = ∛u and we end up with a quadratic equation 6y² + 12y - 48 = 0, which has only two solution for y, namely y = 2 or y = -4, which leads to u = 8 or u = -64. u = 8 and u = x -28 ⇒ x = 36 u = -64 and u = x - 28 ⇒ x = -36. So, the quadratic equation shows we have two solution. Now if the problem didn't have the same coefficients of x under the cube root signs, then there would be three solutions but perhaps not all real and not integers.
Thank you sir
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@@PreMath lñooo9opi8iuuu7jju7jjjjjo8
What so interesting about this that u pinned this comment 🤔🤔
I began college at 51 years of age and cried everyday in algebra. It was a foreign language to me. I'm 70 now and wish I would have knew about you then. I finally quit college after my third year because I felt so dumb 😰. I promised my mom I would graduate. She passed two years ago and I think I can accomplish college once I move out of state. Especially with your lessons.
Best of luck! 👍👍
All the best. Remember to solve as many questions relevant for your exams. You'll get the hang of it eventually.
Do it. I believe You can do it!
good luck to you
We can cube both sides of our equation.
(a-b)^3=a^3-b^3-3ab(a-b)
a=cuberoot(x+28)
b=cuberoot(x-28)
a-b=2
(x+28)-(x-28)-3cuberoot[(x+28)(x-28)]*2=8
56-3(cuberoot(x^2-28^2)*2)=8
48-6(cuberoot(x^2-28^2))=0
8-cuberoot(x^2-28^2)=0
cuberoot(x^2-28^2)=8
x^2-28^2=8^3
x^2-784=512
x^2=1296
x=36 or x=-36
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تمرين جميل . شرح جيد واضح . شكرا جزيلا لكم سيدي . والله يحفظكم ويحميكم ويرعاكم وينصركم جميعها . تحياتنا لكم من غزة غلسطين
very cool problem that turned out a lot simpler than expected. here's another solution:
∛(x+28) - ∛(x-28) = 2
∛(x+28) = 2 + ∛(x-28)
x + 28 = (2 + ∛(x-28))^3
let u = ∛(x-28)
the right hand side then becomes:
(2 + u)^3 = (4 + 4u + u^2)(2+u) = 8 + 4u + 8u + 4u^2 + 2u^2 + u^3 = 8 + 12u + 6u^2 + u^3.
now our equation is
x+28 = 8 + 12u + 6u^2 + u^3
however, u^3 is just x - 28, so we can cancel x from both sides and get
6u^2 + 12u - 48 = 0
which can be factored into 6(u-2)(u+4)
thus, u = 2 or u = -4
substituting u = ∛(x-28):
1: x - 28 = 2^3 ==> x = 36
2: x - 28 = (-4)^3 ==> x = -36
final answer: x = ± 36
2:46 for time limit purposes on tests that don't require solutions, you could do Trial and Error on this point
My first guess was x + 28 = 4³, so x = 36, and realising the equation is "Even" leads x= -36 as the other solution 😂🤣🤣.
I try to guess the solution since the RHS is just 2. may be the radicals are perfect cube roots
1³=1, 2³=8, 3³=27, 4³=64.... Then guess x+28=64 then x=36 substitute that back into the given equation verify x=36 is indeed correct. Since I know we are dealing with cube roots, negative numbers can be solution as well.
∵ (-1)³=-1, (-2)³=-8, (-3)³=-27, (-4)³=-64 ∴ I try to put x=-36 into the given equation and check it out and bingo "yes"
so x can be ±36 x=±36
The solution should've been shorter:
If a = cbrt(x+28), b = cbrt(x-28), and a - b = 2, then
(a-b)^3 = a^3 - b^3 - 3ab(a-b)
8 = 56 - 6ab
ab = 8 --> (ab)^3 = x^2 - 28^2 = 512
x = ±√(512 + 784) = ±36
yes, it's easier and shorter
Let the LHS = f(X), we have f(--X) = f(X), so f is even, we just solve the equation for a positive X and knowing that --X is also a solution.
Hello I was stunned by this problem because I didn’t think that substitution would help at all! In anyway which I would try to solve I would always end up with even more complex equation your way of solving was indeed very need and new! I solved 2nd degree radicals by moving one to the other side then squaring so here I moved to the other side and cubes (cause it’s degree 3) but in the end it would still have cube roots in it thank you for teaching me something new :D
Dear RS, this is a very challenging problem. When I was preparing this problem, I did it many different ways and finally decided to go with the method that I shared in the vid. Thanks dear fro sharing your experience.
You are awesome 👍 Take care dear and stay blessed😃 Kind regards
Love and prayers from Arizona, USA!
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Great , step - by - step solution of a complex solution. It would be difficult to solve without writing out each step to keep track of all the information created by each step. It is also a test of knowledge of binomial expansions and manipulation of information created by each step. Mastery of this skill is a worthy goal. Thank you for the practice!
You have a-b=2 and ab=8
You can arrange a quadratic equation, for instance a^2-2a-8=0 whose solutions are a1=-2 and a2=4
And, of course; b1=-4 b2=2
Using a1=-2 will bring x=36 whereas a2=4 gives the solution x=-36
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Another way would be to solve the equation graphically by y = (3rt x + 28) - (3rt x -28) + 2 and solve when y = 0. Plotting the equation I find with these complex equations is better to understand as it shows it visually. A good advantage of having a graphical calculator
Thanks Daniel for the feedback. You are awesome 👍 Take care dear and stay blessed😃
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Let y=cbrt(x+28)-1, then should be y-1=cbrt(x-28), 56=(x+28)-(x-28)=(y+1)^3-(y-1)^3=6y^2+2, 6y^2=54, y^2=9, y=3 or -3. If y=3, then cbrt(x-28)=2, x-28=8, x=36. If y=-3, then cbrt(x+28)=-2, x+28=-8, x=-36.
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Didn't manage to come up with that substitution at all. Nice!
SIR. AS USUAL YOU ARE FANTASTIC, ADMIRABLE AND MY FAVORITE TUTOR.
MUCH faster if y^3=x-28, then
y^3+56=(y+2)^3, or
y^2+2y-8=0, y:2,-4 which gives
x: 36,-36
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That's exactly what I did. Much faster and easier. Didn't even need to write anything down. There's often some sort of shortcut with these problems.
I applied the cubic formula directly resultantly left with the product of terms under radicals. I took the cube of the equation. My final equation was (x+28)(x-28)=512. Solving for x yielded x=36 and x=-36
Thanks Zaki Naqvi dear for the visit! Thanks for sharing
You are awesome 👍 Take care dear and stay blessed😃 Kind regards
Love and prayers from Arizona, USA!
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Very intereseting solution. Apparently it can be simplified. In step 2 instead of calculating cube of difference, let’s work out square of difference (a-b) ^2=a^2+b^2-2ab=4. Subctract this equation from equation 3 and get ab=8, then multiply a and b: 3sqrt(x^2-784)=8, x=±36.
This is an interesting problem, but the solution has extraneous equations Once you have equations 1 and 3, you can substitute a=b+2 (obtained from equation 1) into the equation a^2+ab+b^2=28. The resulting simplified equation is b^2+2b-8=0, which is quite simple to solve for b.
Awesome question, nice solution! 😊 But here is a faster way: 2 x 28 = 56 = distance between (x + 28) and (x - 28). Because of „2“ on the right site, (x - 28) has to be 8 = 2³. 8 + 56 = 64 = 4³. Therefore: x = 36 (28 + 36 = 64; 36 - 28 = 8). Did it in 2 minutes 😊
a very interesting solution!
Great work. I do not think you need equation 5. Substituting a or b from equation 4 into equation 1 does the same thing. Keep it up! I love your work.
Perfect solving👏👏👏
From a-b=2, b=a-2. Substitute b=a-2 to
a^3-b^3=56, I got a^2-2a-8=0 and got a=-2 or a=4. Finally I got x=-36 or x=36.
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1:46 it would make more sense if you crossed out the index number you know, instead of the top of the radical. That is nothing that needs to be crossed out.
There is a lot of substitution going on here!
This is one of my most favorite math problems out of all the other ones that you upload on TH-cam. I like Logarithms as well! Maybe, you could make up a logarithmic problem that is similar to this! That is my suggestion!
Sure dear
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This is superb solution. I would never have thought of doing it this way.
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Really I enjoyed I left this equation maybe 45 years but you are excellent
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Again, I used a completely different method, but surprisingly got the right answers. I started by moving the second radical to the right side of the equation, then cubed both sides. Then a lot of algebra was required. The key was getting to a quadratic equation in which the variable being squared was a cube root! But from there it worked out very cleanly. I thought my method was cumbersome, but after seeing the official solution, I'm not so sure. I solved for x directly, with no substitution of variables used.
Great job J R. Thanks for sharing your experience. No matter what method you use, it requires extensive algebra.
When I was preparing this problem, I thought about using your method as well. Anyway, thanks for your candid feedback.
You are awesome 👍 Take care dear and stay blessed😃
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I was thought the same way as the person below said about moving the cube root to the other side. This method is a little longer but love it because I have trouble with u substitution in calculus and need more practice. Thank you very much for this video always great to learn a different method, anther tool in the toolbox. 👍
Thanks Adeline for the visit! I put my heart and soul in this video.
You are awesome 👍 Take care dear and stay blessed😃 Kind regards
Love and prayers from Arizona, USA!
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Sir,
With ur substitution we ve, x= 1/2(a^3 + b^3)
a-b=2
a^3 - b^3 =56
Using last two eqns.. We can have a quadratic equation in b. Solve this to get b and hence a. Use these values of a and b in first eqn to get x.
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thankas so much dear!
great solution sir thank u
So nice of you Nico dear! You are awesome 👍 I'm glad you liked it! Please keep sharing premath channel with your family and friends. Take care dear and stay blessed😃
Good video. Great explanation
Write down the first few cube numbers. Find two of them that have a difference of 56. There they are. 64 and 8. Or the corresponding negatives. Check that both pairs work. Done.
That is just guessing and what if the answer was some fraction of what if the answer was a cube or square root too! Then your method wouldn’t work so it’s best to solve systemically
Finding the difference between (x + 28) and (x - 28) is not a guess. It is a calculation. A good mathematician inspects a problem for a feature that might lead to a simple elegant solution. Of course my posting above would need to be expanded into a rigorous solution.
Thank your sir Very Helpful with such good explanation this type of questions are constantly asked in exams.
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Finding value of a,b simply this way .we had a-b=2 and ab=8. So,a+b=under root (a-b)square+4ab which is under rt(4+32)=+-6 giving the value of a,b as shown.
What grade is this please?
Another approach is to take cube on both sides and get the answer in 1 min
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Start with (x+28)^1/3=(x-28)^1/3+2 and take cube of both sides. After simplifications you are left with quadratic equation where (x-28)^1/3 is either 2 or -4.
My math question
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You are the best!!!!!!
Excellent!
You are awesome. Keep it up 👍
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I am new to your channel.
Thanks for sharing this algebraic problem
Keep it up sir.
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Thank you sir, you have helped for me a lot. 🙂🙂🙂🙂
Amazing 🤩
In my view we can also solve this as:-( a+b)^2=(a-b)^2+4ab
Hence we get (a+b) &(a-b), so easily can be solved.
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My math question
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We only need to consider positive x, because if a solution is found to the equation for +x, then -x is also a solution.
As positive x increases, then ∛(x + 28) - ∛(x - 28) is decreasing.
We notice at x = 0, then ∛(x + 28) - ∛(x - 28) = 2∛28 > 2∛27 = 2*3 = 6 and the difference is decreasing towards 0, but is always positive, except at infinite, where the difference is zero.
So, if heuristically we assume that x is a positive integer, then we could try and find a positive cube, say a³, such that x + 28 = a³, where x > 0 to try and find the solution.
The smallest a that meets the above conditions is a = 4 ⇒ a³ = 64 ⇒ x = 36 as a solution, and as -x is also a solution, then x = -36 is another solution.
Have we found all the solutions?
Yes, as ∛(x + 28) - ∛(x - 28) is decreasing and can only be equal to 2 "once", as positive x increases in value.
btw from [1] follows b = 2 -a. Then substitute that into [3]
And also for your nice explanation.
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Wow just amazing 😍
Thanks for your hard work 😸
another way to solve this is just to cube both side or the original equation. (x+28) - (x - 28) - 3 cbrt(x+28) cbrt(x-28) (cbrt(x+28) - cbrt(x-28) ) = 8. which can be simplified as cbrt(x+28) cbrt(x-28) = 8, cube both side again x^2 - 28^2 = 8^3 , and will get x = +/- 36.
Early in the work, you had a-b=2 and ab=8. It would have saved you a lot of work.
Excellent method. Very clear.Thank you
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When I looked at this problem I just didn't know where to start. By considering what the graph of the function looked like I concluded there would be 2 solutions: +X and -X, but more than that I was stuck. Putting the left hand size as a - b in hind sight was obvious but I didn't think of it - once you do it all falls out really nicely.
My math question
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Set u=cube root of (x+28),
v=cube root of (x-28).
Therefore: u^3=x+28, v^3=x-28
So, we have a system of two equations u-v=2 and u^3-v^3=56.
This system of two equation has two solutions, they are u=4, v=2 and u=-2, v=-4.
With u=4, we have x=36,
And with u=-2, we have x=-36.
Summarising, we have two solutions, they are x=36 and x=-36.
Excellent!
@@PreMath thank! I love math. I haven't seen your whole video but a little from the beginning of the video. Thank a lot!
My math question
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Outstanding!
ab=8 means cube roots of (x+28) and (x-28) multiply to get 8. This straight away gives x.
good, but is this just brain exercise or does it have real-world applications?
I solved this a completely different way. Spotted The difference between a and b is 2 and the difference between a^3 and b^3 is 56. Logic said the only way to get these numbers were 2 ,4 and 8,64. Mine was logic - wouldn’t work if it wasn’t an integer problem though. Is my solution valid?
I did the same. LOL Sometimes it's best to take a shot at the easiest solution. Even if it turned out to not be an integer, you'd have a decent bound on the correct answer.
Superb.
En la ecuación 1 despejas a y elevas al cubo, en la segunda despejas a al cubo e igualas.
SUPER!
You r awsome sir
Thank you sir
Thank you for this video 🙂
a^2 + ab + b^2 = a^2 - 2ab + b^2 + 3ab
= ( a - b )^2 + 3ab = 28
= 4 + 3ab = 28
===> ab = 8
Much more simple
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thank you a lot sir appreciate a lot!
Функции f1(x)=cbrt(x+28) и f2(x)=cbrt(x-28) являются обратными от (F(x))^3 функций, f1(x)=f2(x)-2 имеют две точки пересечения. Обозначим t=cbrt(x-28), (t)^3=x-28,
cbrt(x+28)-t=2, cbrt(x+28)=2+t, (cbrt(x+28))^3=(2+t)^3, [х+28=(2+t)^3]-[(t)^3=х-28], х+28-х+28=(2+t)^3-t^3, 56=(2+t-t)((2+t)^2+(2+t)t+t^2, t1^2+2t-8=0, t(1)=2, t(2)=-4, X1=36.
f=cbrt(x+28), f^3=x+28, f-cbrt(x-28)=2, -cbrt(x-28)=2-f, (cbrt(x-28))^3=(f-2)^3, [x-28=(f-2)^3]-[f^3=x+28], x-28-x-28=(f-2)^3-f^3, -56=(f-2-f)((f-2)^2+(f-2)f+f^2, t1^2-2t-8=0, f(1)=-2, f(2)=4, X2=-36.
Ótima equação irracional. Resolução muito interessante. Não deixa de ser um bom desafio👍👍👍👍
starting from equation 4 and cubical it makes x^2 -28^2=4^3 and we can work out X
Great!
a+b = 6 or -6
Had to do some thought in order so solve it, but I got it
Awesome puzzle, like always ❤️
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My math question
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a^3-3a^2b+3ab^2-b^3
But your approach was really great! Thanks for this question.
I have solved it another approach. Let's see.
(x+28)^(1/3) - (x-28)^(1/3)=2
Or, (x+28) - (x-28) - 3{(x+28) (x-28)}^(1/3) {(x+28)^(1/3) - (x-28)^(1/3)=8 [cubing both].
Or, 56 - 3{ x^2 - (28)^2}^(1/3) * 2 = 8 [ because (x+28)^(1/3) - (x-28)^(1/3) =2 have given]
Or, 6(x^2 - 784)^(1/3) = 48
Or, (x^2 - 784)^(1/3) = 8
Or, x^2 -784 = 512 [cubing both again]
Or, x^2 = 1296
So, x = plus minus 36 (+ or - 36)
Why don't you just isolate either a or b then substitute it into 2 then expand using the binomial theorem and cancel the cubic terms?
No matter what method you use, it requires extensive algebra.
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Primer in the wall.
Don’t need second substitution just write x in terms of a
easy enought, the cbrt of 32 can be broken to cbrt(8*4) then we have 2 * cbrt( 8 * 8 * 4 /2) which is 2*2*2*cbrt(2) which is 8 cbrt(2)
Are we talking about principal roots or real roots? Generally, if not specified it‘s the principle value. But then, cbrt(-8) is not -2.
Looking at the problem, i know the solution is long. Those who dislikes math hates long solution. Others just don't know where to start. Math is practice and those who are good in math have their own method in solving math problems. Having said that there are more than one way to solve a math problem. Personally, my method is automatic the moment I see a math problem. In worded problems though, the first 4 steps are write the given, then what the problem is asking, then draw a diagram or figure and fourth the solution. The solution is always base on your analysis of the figure or diagram.
Let u = x - 28, so u + 56 = x + 28
Place these in the original equation and you get
∛(u + 56) - ∛u = 2
⇒ ∛(u + 56) = 2 + ∛u
⇒ u + 56 = (2 + ∛u)³ =
8 + 3*2² * ∛u + 3*2¹ * (∛u)² + (∛u)³ =
8 + 12∛u + 6(∛u)² + u
⇒ 6(∛u)² + 12∛u - 48 = 0
Let y = ∛u and we end up with a quadratic equation
6y² + 12y - 48 = 0, which has only two solution for y,
namely y = 2 or y = -4, which leads to u = 8 or u = -64.
u = 8 and u = x -28 ⇒ x = 36
u = -64 and u = x - 28 ⇒ x = -36.
So, the quadratic equation shows we have two solution.
Now if the problem didn't have the same coefficients of x under the cube root signs, then there would be three solutions but perhaps not all real and not integers.
sir, It's simpler if you just use equation 1 an 2. I arrived at the same answers.
Surprisingly complex, but well explained and an interesting problem!
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A very wierd function to work with... I let f(x)=(x+28)^(1/3) and g(x)=(x-28)^(1/3)... then carried out with the pattern of which f(x)-g(x)=2. I recirved 2 vertical lines; x= 36 and x=-36. 🤔
On 6:43, systems Eq(1) and Eq (4), you can solve it…
Ok, that was a bit harder than I thought.
So I rearrange:
sq3(x+28) = 2 + sq3(x-28)
I raise both sides to the cube :
x+28 = 8 + 3*4* sq3(x-28) + 3*2*sq3(x-28)² + (x-28)
So
48 = 3*4* sq3(x-28) + 3*2*sq3(x-28)²
8 = 2* sq3(x-28) + sq3(x-28)²
I pose y = sq3(x-28) and I get
y²+2y-8 = 0
So y = 2 or y = -4
As x-28 is negative, y must be negative too, so y = -4 ( Edit: Wrong assumption, see answer below)
so y == sq3(x-28) = -4
So I raise to the cube and I get:
x-28 = -64
x = -36
Let's check
The initial expression evaluates to
sq3(-36+28) - sq3(-36-28) = sq3(-8) - sq3(-64) = 2 - (-4) = 2
Which confirms my solution.
Ah, it seems I missed the other solution which is x=36
I'm not sure where I lost that solution along the way.
Oh yes, I can't just say that x-28 is negative...
So I should test y = 2
Which give
x-28 = 8
So x = 36
Somehow I find my solution easier to follow but that is probably just my brain being biased in favor of itself.
It' enough a, or b for solution.
you solved this easily, very well done
We have two solutions.
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awesome!!!
a^2+b^2=20 is unnecessary. Because from (3), u can easily get (a+b)^2-ab=28, and since ab=8, therefore, a+b=6.
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Superb, as always
Thanks again!
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Guruji,pranam. The answer is x= +36 and -36.
Great job Jaggi dear
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x = 36 or -36
a^(2) + b^(2) = 20
I want to send a simple solution without taking substitutes. How can I contact you.
Dear Zaki, email us at premathchannel@gmail.com
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100 th like
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In H.S.L.C final exam that is class X (Assam, India) 1983 similar question was given with 2 marks. Unfortunately, I failed to solve. Thanks a lot for your maths classes.
My math question
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Could someone explain why there is only 2 solutions and not 3?
Let u = x - 28, so u + 56 = x + 28
Place these in the original equation and you get
∛(u + 56) - ∛u = 2
⇒ ∛(u + 56) = 2 + ∛u
⇒ u + 56 = (2 + ∛u)³ =
8 + 3*2² * ∛u + 3*2¹ * (∛u)² + (∛u)³ =
8 + 12∛u + 6(∛u)² + u
⇒ 6(∛u)² + 12∛u - 48 = 0
Let y = ∛u and we end up with a quadratic equation
6y² + 12y - 48 = 0, which has only two solution for y, namely y = 2 or y = -4, which leads to u = 8 or u = -64.
u = 8 and u = x -28 ⇒ x = 36
u = -64 and u = x - 28 ⇒ x = -36.
So, the quadratic equation shows we have two solution.
Now if the problem didn't have the same coefficients of x under the cube root signs, then there would be three solutions but perhaps not all real and not integers.
I have a TI inspire CD CAS and got
X=36 or
X=-36.0000000001
Why the 0000000001 at the end ???
x=36
Great job my friend
We have two solutions.
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