Bonjour, Vous commencez votre démonstration avec l’idée d’utiliser le 8, parce que 8=2^3. Alors, si on utilise son intuition pour commencer, puis son expérience , pour « voir » à 5:40 , que a =b=1, on peut aussi se permettre de dire que le « squ5 » donne l’idée d’utiliser le nombre d’or. Il vérifie l’égalité x^2=x+1, donc. x^3 = x^2 + 1, puis. x^3 = (x + 1) + x . Avec x=(1+squ5) / 2, on a 2x+1= 2 + squ5. C’est donc OK Mon côté cartésien ( bien français ?), me fait penser que ce « problème » était plutôt une très belle devinette, et ça m’a fait passer un très bon moment. Merci! PS desolee d’écrire en français, mais mon anglais est tres mauvais !!!!
I guess the unstated issue is what the word "simplify" implies in the question. It means can I find an expression without nesting of root signs in it? Suppose the answer was something like ( a + b root c + d root e )? Well, firstly, there couldn't be lots of terms like root c and root e and so forth without getting terms with root c root e when you cubed it, so it must be only two terms: ( a + b root c ). Secondly, the c would have to be 5, because the root 5 term must come from the second and fourth terms of the expansion. And, thirdly, the a and b can't be fractions with large denominators, because those denominators would quickly get huge when the answer was cubed. It won't be "simplified" unless the answer is in the form a + b root 5 for some fairly small values of a and b.
Thank you for this nice, little problem. To be honest, i was unable to solve it. But if some guesswork is allowed, you can try directly (u+v√5)³ = 2 + √5 with natural numbers u and v. Now one finds with your formular (a + b)³ = a³ + 3a²b + 3ab² + b³: (u + v√5)³ = u³ + 3v√5u² + 3 v²√5²u + √5³v³ = u³ + 3vu²√5 + 15v²u + 5v³√5 = (u³ + 15v²u) + (3vu² + 5v³)√5 := 2 + 1√5 comparison results to another guess: (1) u³ + 15v²u = 2 (2) 3vu² + 5v³ = 1 Now to the smart part: the sum of coefficients on the left hand side in (1) is 1 + 15 = 16 doubles the similar sum in (2): 3 + 5 = 8, and the right hand side of (1) also double the right hand side in (2). This lets us try the final guess: v = u. Inserting this equality into (1) and (2) leads to (1´) u³ + 15 v²u = u³ + 15 u²*u = u³ + 15u³ = (15 + 1) u³ = 16u³ = 2 => u³ = 1/8 => u = v = 1/2 (2´) 3vu² + 5v³ = 3 u*u² + 5 u³ = 3u³ + 5 u³ = (3 + 5) u³ = 8u³ = 1 => ... => u = v = 1/2 inserting this into the given problem leads to the same final as your aproach: (2 + √5)^(1/3) = (1/2 + √5/2)³^(1/3) = 1/2 (1 + √5). Of cause, you should check this by taking to the cube: (1/2 (1 + √5))³ = 1/8 * (1+√5)³ = 1/8 * (1³ + 3 * 1² * √5 + 3 * 1 * √5² + √5³)= 1/8 * (1+ 3*√5³ + 3 * 5 + 5√5) = 1/8 * (16+8√5) = 2 +√5 as expected.
Alternate method of solution: make a new equation, cubert(2 + sqrt5) + cubert(2 - sqrt5) = c solve for c by cubing both sides: 4 - 3c = ccc You can solve the cubic equation. You get 3 roots, but 2 of them are imaginary. And you know that cubert(2 + sqrt5) must be a real number because cuberoot of a real number gives a real number. So you can eliminate the imaginary roots as solutions once you have a value for c (it turns out, c = 1) you have: cubert(2 + sqrt5) + cubert(2 - sqrt5) = 1 since you want the value of cubert(2 + sqrt5), set x = cubert(2 + sqrt5) x + cubert(2 - sqrt5) = 1 cubert(2 - sqrt5) = 1 - x cube both sides 2 - sqrt5 = 1 - 3x + 3xx - xxx substitute xxx = 2 + sqrt5 2 - sqrt5 = 1 - 3x + 3xx - (2 + sqrt5) and from here, you have a quadratic equation in terms of x that is easy to solve
The shorthand for cube root is "cbrt," just in the same way that "sqrt" is the shorthand for square root. So, your whole post should have "cbrt" written in place wherever you have "cubert" written.
@@strikerstone It's a lot easier to solve a single cubic equation with one variable (4-3c=ccc) than it is to solve a system of two cubic equations with two variables.
@@strikerstone Sure. let's say: u = 2 + sqrt5 v = 2 - sqrt5 u + v = 4 u * v = 4 - 5 = -1 cuberoot(u) + cuberoot(v) = c cube both sides u + 3cuberoot(uv)(cuberoot(u) + cuberoot(v)) + v = c^3 we can substitute -1 for uv and we can substitute c for cuberoot(u) + cuberoot(v) to get: u + v + 3cuberoot(-1)c = c^3 we know cuberoot(-1) = -1 and we can substitute u+v = 4 and move everything to the same side c^3 + 3c - 4 = 0 this is a cubic equation. By the rational root theorem, if there is any integer solution, it would be one of the following: -4, -2, -1, 1, 2, 4 so it makes sense to test those 6 values first if you test them all, you will see that c =1 is a solution. if none of those 6 values worked, that would mean the answer is not rational. There are other ways to solve a cubic equation, namely the cubic formula Since you have a solution (c=1) you can factor out (c-1) from the equation in order to find the other 2 solutions you get that c^3 + 3c - 4 = (c-1)(c^2 + c + 4) then you can use the quadratic formula to solve c^2 + c + 4 = 0. You will see the two solutions are both complex. And going back to the beginning of the problem cuberoot(2 + sqrt5) and cuberoot(2 - sqrt5) are both real numbers, not complex. So of the 3 solutions to the cubic equation, you reject the two complex ones, and you are left with "1"
@6:15 Assuming a and b are integers, then we should really factorise a³ + 15ab² = 16 as a(a² + 15b²) = 16. So, a could be equal to 1, 2, 4, 8 or 16, as these are the factors of 16 and a must be positive because a² + 15b² is positive. However, if we try these values of a, then only a = 1 is a possible solution with a² +15b² = 16 if b is to be an integer, requiring that b² = 1, so b = -1 or 1. However if we try b = -1 with a = 1 in 3a²b + 5b³, we get -3 - 5 = - 8 ≠ 8, so this is not a solution. However, when we try a = 1 and b = 1 in 3a²b + 5b³, we get 3 + 5 = 8, which is a solution. So, (a, b) = (1, 1) is the *only * integer solution possible.
that's what i say all those formulas are not teaching how to get to the answer 1.61803399 plus they are vague questions it's all abstractions which is the worst way to teach math
Actually (2+√5) is the outcome of 1/8(1+√5) power3 .that is why 8,/8 multiplication is applied.if you work out (1+√5)power 3, you will get 8(2+√5).students should keep it in mind like other formulas.
It was clear from the onset that the expression inside the root had to be a perfect cube, the issue was to find the path to find the solution. And I wonder if there is still another pair of _a_ and _b_ that satisfies the equation.
Uses the same method as described in the video, we can deduce that ∛(2 - √5) = (1 - √5)/2 So, ∛(2 + √5) + ∛(2 - √5) = (1 + √5)/2 + (1 - √5)/2 = 1. In fact, if we use the cubic equation formula to find the real root of x³ + 3x - 4 = 0, we find the answer is x = ∛(2 + √5) + ∛(2 - √5), which by inspection of the cubic equation requires the real root of x to be x = 1.
Cuberoot of (2+√5) is the given problem We know that universal formula Cube (a+b)= Cube a+ Cube b+ 3ab(a+b) Cuberoot of cube (a+b) is the given problem. It can be written as below Cube of (2+√5)=Cube a+ Cube b+ 3 ×2×√5(2+√5) Compare this with Cube a+ cube b+3ab(a+b)=Cube(a+b) Hence, Cuberoot of Cube (2+√5)=2+√5 (2+√5) is the answer . Basically , I am 62yrs old physician practicing medicine. I am interested in mathematics from my school days. Out of curiosity I am solving the mathematical problem. If I am wrong, please forgive me
Hi, I have a math question about The perimeter of an isosceles triangle is 38 m. Find the length of the sides of the triangle of maximum area. Would you be able to make a video and explain it?
Heron képletből kell kiindulni, melyekben van: a, c=b, majd kapok egy függvényt, ebben, b=(38-a)/2, ezután x=a. A terület az "y". Majd deriválással megkeresem a maximumot.
@@zerilioner639 Draw an isosceles triangle. Label the two equal sides "a". Label the 3rd side "b". Then draw a perpendicular line from side "b" to the opposite vertex. Label this side "h". Area of the triangle is (1/2)*b*h. Now you need to get "b" and "h" in terms of a single variable. "a" is the easiest. Perimeter of the triangle is 38. So a + a + b = 38, or 2a + b = 38. So therefore b = 38-a. Area of the triangle is now (1/2)*(38-a)*h. The perpendicular line from "b" to the vertex bisects b. So you have a right triangle with sides (b/2) and (h) and hypotenuse (a). So (b/2)^2 + h^2 = a^2. Solve for h, plug in to the Area equation. Take the derivative of the Area equation with respect to "a", set result equal to zero. Solve for "a". That's the "a" that maximizes the value of the area.
I wrote it as x = 3√(2+√5). Then, cubing both sides, x^3 = 2+√5. Then squaring both sides, x^6 = 4 + 4√5 + 5, simplifying to x^6 = 9 + √(16x5), simplifying to x^6 = 9 + √80, then used a calculator to yield x=1.618... WHICH calculates out to be THE SAME NUMBER AS HE GOT!
@@doubledee9675 well yes the golden ratio is the same as the original but written much simpler, since there is only one square root instead of a square and a cube root. You cant expect this thing to magically turn into a 1 or a 2 you know
Amazing problem ... Millions of thankuuu sir 💓💓
Thanks Sangam for the visit! You are awesome 👍 Take care dear and stay blessed😃 Kind regards
@@PreMath Isn't this answer the *golden ratio* ? 😱😱
The occurrence of introducing 8/8 is simply Amazing. Don't think it will occur to anyone easily.
Excellent. Hats off to the Professor.🙏
Great , but is there any short trick or general formula to find it out ?
Well, you can easily manage to turn such numbers to a square, but a cube would be a bit more difficult.
Bonjour,
Vous commencez votre démonstration avec l’idée d’utiliser le 8, parce que 8=2^3.
Alors, si on utilise son intuition pour commencer, puis son expérience , pour « voir » à 5:40 , que a =b=1, on peut aussi se permettre de dire
que le « squ5 » donne l’idée d’utiliser le nombre d’or. Il vérifie l’égalité x^2=x+1, donc. x^3 = x^2 + 1, puis. x^3 = (x + 1) + x .
Avec x=(1+squ5) / 2, on a 2x+1= 2 + squ5. C’est donc OK
Mon côté cartésien ( bien français ?), me fait penser que ce « problème » était plutôt une très belle devinette, et ça m’a fait passer un très bon moment. Merci!
PS desolee d’écrire en français, mais mon anglais est tres mauvais !!!!
Thank you, Sir! Looked tough but after your excellent explanation, the problem becomes easy.
I guess the unstated issue is what the word "simplify" implies in the question. It means can I find an expression without nesting of root signs in it? Suppose the answer was something like ( a + b root c + d root e )? Well, firstly, there couldn't be lots of terms like root c and root e and so forth without getting terms with root c root e when you cubed it, so it must be only two terms: ( a + b root c ). Secondly, the c would have to be 5, because the root 5 term must come from the second and fourth terms of the expansion. And, thirdly, the a and b can't be fractions with large denominators, because those denominators would quickly get huge when the answer was cubed. It won't be "simplified" unless the answer is in the form a + b root 5 for some fairly small values of a and b.
Thank you for this nice, little problem. To be honest, i was unable to solve it. But if some guesswork is allowed, you can try directly (u+v√5)³ = 2 + √5 with natural numbers u and v.
Now one finds with your formular (a + b)³ = a³ + 3a²b + 3ab² + b³:
(u + v√5)³ = u³ + 3v√5u² + 3 v²√5²u + √5³v³ = u³ + 3vu²√5 + 15v²u + 5v³√5 = (u³ + 15v²u) + (3vu² + 5v³)√5 := 2 + 1√5
comparison results to another guess:
(1) u³ + 15v²u = 2
(2) 3vu² + 5v³ = 1
Now to the smart part: the sum of coefficients on the left hand side in (1) is 1 + 15 = 16 doubles the similar sum in (2): 3 + 5 = 8, and the right hand side of (1) also double the right hand side in (2). This lets us try the final guess: v = u. Inserting this equality into (1) and (2) leads to
(1´) u³ + 15 v²u = u³ + 15 u²*u = u³ + 15u³ = (15 + 1) u³ = 16u³ = 2 => u³ = 1/8 => u = v = 1/2
(2´) 3vu² + 5v³ = 3 u*u² + 5 u³ = 3u³ + 5 u³ = (3 + 5) u³ = 8u³ = 1 => ... => u = v = 1/2
inserting this into the given problem leads to the same final as your aproach: (2 + √5)^(1/3) = (1/2 + √5/2)³^(1/3) = 1/2 (1 + √5).
Of cause, you should check this by taking to the cube:
(1/2 (1 + √5))³ = 1/8 * (1+√5)³ = 1/8 * (1³ + 3 * 1² * √5 + 3 * 1 * √5² + √5³)= 1/8 * (1+ 3*√5³ + 3 * 5 + 5√5) = 1/8 * (16+8√5) = 2 +√5
as expected.
Alternate method of solution:
make a new equation, cubert(2 + sqrt5) + cubert(2 - sqrt5) = c
solve for c by cubing both sides: 4 - 3c = ccc
You can solve the cubic equation. You get 3 roots, but 2 of them are imaginary. And you know that cubert(2 + sqrt5) must be a real number because cuberoot of a real number gives a real number. So you can eliminate the imaginary roots as solutions
once you have a value for c (it turns out, c = 1) you have:
cubert(2 + sqrt5) + cubert(2 - sqrt5) = 1
since you want the value of cubert(2 + sqrt5), set x = cubert(2 + sqrt5)
x + cubert(2 - sqrt5) = 1
cubert(2 - sqrt5) = 1
- x
cube both sides
2 - sqrt5 = 1 - 3x + 3xx - xxx
substitute xxx = 2 + sqrt5
2 - sqrt5 = 1 - 3x + 3xx - (2 + sqrt5)
and from here, you have a quadratic equation in terms of x that is easy to solve
The shorthand for cube root is "cbrt," just in the same way that "sqrt" is the shorthand for
square root. So, your whole post should have "cbrt" written in place wherever you have "cubert"
written.
this is shit , we have to do the same thing what we are doing here to find your C , how you got 1 bruh?
@@strikerstone It's a lot easier to solve a single cubic equation with one variable (4-3c=ccc) than it is to solve a system of two cubic equations with two variables.
@@armacham can you give me a detailed explanation from (2+√5)^1/3 + (2-√5)^1/3 = c , to How to got 1?
@@strikerstone Sure.
let's say:
u = 2 + sqrt5
v = 2 - sqrt5
u + v = 4
u * v = 4 - 5 = -1
cuberoot(u) + cuberoot(v) = c
cube both sides
u + 3cuberoot(uv)(cuberoot(u) + cuberoot(v)) + v = c^3
we can substitute -1 for uv
and we can substitute c for cuberoot(u) + cuberoot(v)
to get:
u + v + 3cuberoot(-1)c = c^3
we know cuberoot(-1) = -1
and we can substitute u+v = 4
and move everything to the same side
c^3 + 3c - 4 = 0
this is a cubic equation. By the rational root theorem, if there is any integer solution, it would be one of the following:
-4, -2, -1, 1, 2, 4
so it makes sense to test those 6 values first
if you test them all, you will see that c =1 is a solution.
if none of those 6 values worked, that would mean the answer is not rational. There are other ways to solve a cubic equation, namely the cubic formula
Since you have a solution (c=1) you can factor out (c-1) from the equation in order to find the other 2 solutions
you get that c^3 + 3c - 4 = (c-1)(c^2 + c + 4)
then you can use the quadratic formula to solve c^2 + c + 4 = 0. You will see the two solutions are both complex.
And going back to the beginning of the problem cuberoot(2 + sqrt5) and cuberoot(2 - sqrt5) are both real numbers, not complex. So of the 3 solutions to the cubic equation, you reject the two complex ones, and you are left with "1"
@6:15
Assuming a and b are integers, then
we should really factorise a³ + 15ab² = 16 as
a(a² + 15b²) = 16.
So, a could be equal to 1, 2, 4, 8 or 16, as these are the factors of 16 and a must be positive because a² + 15b² is positive.
However, if we try these values of a, then only a = 1 is a possible solution with a² +15b² = 16 if b is to be an integer, requiring that b² = 1, so b = -1 or 1.
However if we try b = -1 with a = 1 in 3a²b + 5b³, we get -3 - 5 = - 8 ≠ 8, so this is not a solution.
However, when we try a = 1 and b = 1 in 3a²b + 5b³, we get 3 + 5 = 8, which is a solution.
So, (a, b) = (1, 1) is the *only * integer solution possible.
मस्त,आवडलं
the value of ³√(2+√5) is approximately 1.62
all numbers around 1.62 are (1+√5)/2.
that's what i say
all those formulas are not teaching how to get to the answer
1.61803399
plus they are vague questions
it's all abstractions which is the worst way to teach math
Thank you very much sir 🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻
Greetings from the BIG SKY. Good problem.
Actually (2+√5) is the outcome of 1/8(1+√5) power3 .that is why 8,/8 multiplication is applied.if you work out (1+√5)power 3, you will get 8(2+√5).students should keep it in mind like other formulas.
Why did you rise the numerator to the power 3 except the dinomerator
very good.
thank you teatcher.
You are a very good teacher.
Thank you.
You are welcome dear!
Thanks Elisheva for the feedback. You are awesome 👍 Take care dear and stay blessed😃
Hello my friend! A very smiliar question with the one I covered not too long ago~ Haha you did a fantastic job as always! Lets keep rocking!
It was clear from the onset that the expression inside the root had to be a perfect cube, the issue was to find the path to find the solution.
And I wonder if there is still another pair of _a_ and _b_ that satisfies the equation.
No. Although the presenter did not show it, the only integer solution is (a, b) = (1, 1).
Thanks for nice solving. It was really great.
Most welcome dear 😊
Thanks Rahman dear for the feedback. You are awesome 👍 Take care dear and stay blessed😃
Uses the same method as described in the video, we can deduce that ∛(2 - √5) = (1 - √5)/2
So, ∛(2 + √5) + ∛(2 - √5) =
(1 + √5)/2 + (1 - √5)/2 = 1.
In fact, if we use the cubic equation formula to find the real root of x³ + 3x - 4 = 0, we find the answer is
x = ∛(2 + √5) + ∛(2 - √5), which by inspection of the cubic equation requires the real root of x to be x = 1.
excellent! thank you sir!
Could you explain how and why you put = (a+b sqrt5)^3 / 2^3? Why is that equal to (16 + 8 sqrt5) / 8? Thank you for the video
I was wondering that too!
Could it be because we want the top and bottom to be cubed, since we shall find the cube root, which will cancel the cubes?
beautiful solution sir thank u and keep it up!
Very good.
nice
You are genius.
HITH do figure the new result is simplified???
Because there is only one radical, and it is lower order.
What a creative solution!
Thanks dear Minh
You are awesome 👍 Take care dear and stay blessed😃
The answer (1 + Sqrt(5)) / 2 = Phi, the golden ratio. Phi^3 = 2 + Sqrt(5).
Yo long time dont see, im 5plus5i, remember me?
amazing!!!
why the 8 i understand that we need a cube root but why 2
I feel like that step comes too early, you’d first inspect the coefficients of the expanded cube and see what you need - and only 8 does the job.
Cuberoot of (2+√5) is the given problem
We know that universal formula
Cube (a+b)= Cube a+ Cube b+ 3ab(a+b)
Cuberoot of cube (a+b) is the given problem. It can be written as below
Cube of (2+√5)=Cube a+ Cube b+ 3 ×2×√5(2+√5)
Compare this with
Cube a+ cube b+3ab(a+b)=Cube(a+b)
Hence,
Cuberoot of Cube (2+√5)=2+√5
(2+√5) is the answer .
Basically , I am 62yrs old physician practicing medicine. I am interested in mathematics from my school days. Out of curiosity I am solving the mathematical problem. If I am wrong, please forgive me
Have you ended up where you started?
Very helpful
By observing that (1+✔5)^3=16+8✔5 we can short cut the steps to solve the problem.
great videos sir! I leaned a lot from PreMath.
Great to hear! You are too generous my friend.
Thanks dear for the feedback. You are awesome 👍 Take care dear and stay blessed😃
very well done, thanks for sharing
Thank you! Cheers!
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o my goodnessss!!!!
My issue in this problem is that it can be solve by a scientific calculator. There is no unknown inside the cube root.
Very nice solution 👍
Let x=(2+√5)^1/3 ; -1/x=(2-√5)^1/3
And we know that x-1/x=1
So x=(√5+1)/2 ans
Thank you.
btw the answer at last wast surprisingly the golden ratio
The Golden ratio!
Wow
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Thank u sir . I am from nepal
Ей-богу, я ни за что бы не решил так лихо!
golden ratio in answer !
Why does (16+8√5)/2^3 = (a+b√5)^3 / 2^3?
This is the step/leap in the process that does not make sense.
mindblown! :D
Golden ratio
Hi, I have a math question about The perimeter of an isosceles triangle is 38 m. Find the length of the sides of the triangle of maximum area. Would you be able to make a video and explain it?
Heron képletből kell kiindulni, melyekben van: a, c=b, majd kapok egy függvényt, ebben, b=(38-a)/2, ezután x=a. A terület az "y". Majd deriválással megkeresem a maximumot.
Easy if you know a little calculus. Not quite as easy if you don't.
@@highlyeducatedtrucker Easy? Please let me know the solution.
@@zerilioner639 Draw an isosceles triangle. Label the two equal sides "a". Label the 3rd side "b".
Then draw a perpendicular line from side "b" to the opposite vertex. Label this side "h".
Area of the triangle is (1/2)*b*h.
Now you need to get "b" and "h" in terms of a single variable. "a" is the easiest.
Perimeter of the triangle is 38. So a + a + b = 38, or 2a + b = 38. So therefore b = 38-a.
Area of the triangle is now (1/2)*(38-a)*h.
The perpendicular line from "b" to the vertex bisects b. So you have a right triangle with sides (b/2) and (h) and hypotenuse (a). So (b/2)^2 + h^2 = a^2.
Solve for h, plug in to the Area equation. Take the derivative of the Area equation with respect to "a", set result equal to zero. Solve for "a". That's the "a" that maximizes the value of the area.
@@highlyeducatedtrucker Can you be able to make a video to explain it clearly?
7:45 Yes, this is better. The way you cancelled what needs to undo.
Thanks Aakash for the visit! You are awesome 👍 Take care dear and stay blessed😃 Kind regards
@@PreMath You're Welcome! You too stay blessed.
Golden ratio!!
Excellent Jay
Thanks dear for the feedback. You are awesome 👍 Take care dear and stay blessed😃
Earlier Last step not clearly disscus.....
Guessing
👍
Thanks German for the visit! You are awesome 👍 Take care dear and stay blessed😃 Kind regards
@@PreMath Й тобі дякую. Я не знаю англійської, хочу вивчити, тому пізніше перекладу.
I wrote it as x = 3√(2+√5). Then, cubing both sides, x^3 = 2+√5. Then squaring both sides, x^6 = 4 + 4√5 + 5, simplifying to x^6 = 9 + √(16x5), simplifying to x^6 = 9 + √80, then used a calculator to yield x=1.618... WHICH calculates out to be THE SAME NUMBER AS HE GOT!
(1+sqrt5)/2
5 root 9
I'm not sure how that answer is any simpler than the starting point.
the answer is the golden ratio
@@Livio_05 Yes, but then (i) that's not exactly a simplified concept and (ii) is it a simplification of the original?
@@doubledee9675 well yes the golden ratio is the same as the original but written much simpler, since there is only one square root instead of a square and a cube root. You cant expect this thing to magically turn into a 1 or a 2 you know
@@Livio_05 I was hoping for that.
Answer is the golden ratio
(1,1)
Хочу попросить автора писать цифры большим шрифтом, а то плохо смотреть видео, очень мелкие.
Спасибо за то, что дали нам знать.
Большое спасибо за отзыв. Ты потрясающий 👍 Береги себя, дорогая, и оставайся счастливым
The problem is not challenging and not interesting
And your answer is a famous number; it’s phi
Wow!
Thanks Denis for the feedback. You are awesome 👍 Take care dear and stay blessed😃
How did he get 8 ➗ 8
Try simplifying ∛(3 + √5) using the same method 😀.