Split the quotient: 27^(1/3) * (1/16)^(1/3). 16 can be written as: 2 * 8. That means: 3 * (1/(2 * 8))^(1/3). 2^3 = 8. That means that the cube root of 8 is 2. So, the equation can also be written like: 3 / (2 * 2^(1/3)). Or: (3/2) / (2^(1/3)). Or: 1.5 * 2^(-1/3).
i just came across this and it has been many years since doing this kind of math. this is what i came up with: step one- multiply by cube root of 16 / cube root of 16 to get cube root of 27 times cube root of 16 /16; step two- simplify to 3 times cube root of 8 times cube root of 2 /16; step three- simplify to 3 times 2 times cube root of 2 / 2 times 8; step four simplify to 3 times cube root of 2 / 8. where and why did i go wrong?
The problem is in the first step. In the denominator, cube root of 16 * cube root of 16 is not 16. Almost did this myself because I am used to having a square root in the denominator, so just multiply by itself to get rid of the radical.
You all can look at it this way. When you have an a^(1/r) in the denominator, you need to give this term r-1 identical "buddies" and then put a copy of each of those "buddies" in the numerator to work toward the proper radical format when presenting a value as an answer. If it is a square root, you need one (2 - 1) "a^(1/2)" in both the numerator and denominator. if it were a seventh root then you, of course, would need six (7 - 1) "a^(1/7)"s in both the numerator and denominator.
I would have first simplified it as: 3*cube_root(1/16) using the multiplication property of radicals, reading 27/16 as 27 * 1/16. Then you can reduce cube_root(1/16) by using the equivalent fraction 4/64, and the division property of radicals to get 3*cube_root(4)/4
The way I've seen it in classes and on these math channels, is they tend to frown on having fractional exponents in the final answer, or exponents in the radicand, unless logistically necessary (ie: if it's more feasible to express the radicand as an exponentiated term rather than a rational number). Thus ∛4 is more ideal than 2^(2/3) or ∛2², but ∜71³ would be preferable to ∜357911.
@@jamesharmon4994 Why? I have run into many equations used in engineering and science that have radicals in the denominator. If it's good enough to design bridges and spacecraft, it's good enough for any Real World application.
It would make more sense to, in the last step, multiply by the cube root of 2 over the cube root of 2 twice...effectively cubing the cube root of 2 in the denominator to give 2; then simplifying the 2 cube roots of 2 in the numerator to the cube root of 4.
I got part way through doing this in my head and saw your cube root of 4 in the answer and said, "What the ...???" I was thinking of just multiplying the cube root of 2 times the cube root of 2. But I was wrong, as you showed. Doing that will still leave a radical in the denominator. Oops! Good problem and good explanation. But please try not to repeat yourself so much. You explained the 7/√3 example three times.
I didn't know how to rationalize a cube root denominator before watch your solution I was looking for ways to solve it this is what I thought the cube root of 16 is 16^1/3 so if I times both numerator and denominator by 16^2/3 it would leave 16^3/3 so just 16 3x 16^2/3 = 3 x cube root 256 3 x cube root of 64x4 3x4x cube of 4 / 16 3*Cube root 4 / 4 Your way is much easier
Well explained! I got to 3/the cube root of 16 but I knew it was not enough. Repeat after me, we can't have an irrational number as the denominator. :( Got it! Thanks!
I didn't get it right at first, but I understand your explanation. I am a 71-year-old senior citizen student at my local community college, with a 3.9 GPA. 😊
My math OCD doesn't like cubes in the numerator. I'd still like to solve it, as it seems like it doesn't have a "true" final answer. It would have a decimal that never ends, most likely. 🤔
If your're trying to write 3/2*cuberoot(2), thats correct but you have to rationalize the denominator. To do that we multiply top and bottom by cuberoot(2)^2. cuberoots must be multiplied out 3 times unlike square roots. So apply that to the denominator, you get 2cuberoot(2) * cuberoot(2)^3 which gives 4 and on top multiply 3*cuberoot(2)^2, we square it here because there was not a cuberoot(2) already there like in the denominator. this gives 3cuberoot(4)/2
its (3*cuberoot(4))/4. What you have is correct but you have to rationalize the denominator. Since we are dealing with cuberoots we have to multiply top and bottom by cuberoot(2) twice (cuberoot(2)*cuberoot(2)*cuberoot(2))=2 and on top you have 3*(cuberoot(2))^2 = (3cuberoot(4)/4)
Disagree. Acceptable answer is 3/(2*cube root(2)). Reason, Sin 45 is 1/sqrt(2) not sqrt(2)/2. It is acceptable to have radicals in the denominator, well, it was when I was at secondary school 55 years ago.
For this comment, I’ll not even try to solve the question. Just a thought. ‘Many will get wrong’. In all vids. True statement, but it is phrased in the negative, missing an ‘it’ I guess on purpose, some psychology coming in… Personally, not a creator on TH-cam , I’d go with ‘Did you get it right?’. Feels much more engaging to me. Just my thoughts, let the shredding begin, as long as it stays within the boundaries of math.
I'm getting (3*³√2)/4 oops should have multiplied ³√4/³√4 for my "one" used ³√2/³√2 instead (thinking squares instead of cubes. Correct answer is (3*³√4)/4.😂
Super simple I will use fraction exponents to make it easier to read. Numerator first: 27 ^ 1/3 = (3 * 3 * 3) ^ 1/3 = 3 This shows that 27 is a perfect Cube Denominator: 16 ^ 1/3 = (2 * 2 * 2 * 2) ^ 1/3 = 2(2^1/3) This shows 16 is NOT a perfect cube but since 8 is we can rewrite 16^1/3 as (8 * 2) ^ 1/3. Which gives me 2(2^ 1/3) 2(2 * 4) ^ 1/3 = 2(8 ^ 1/3) = 2 * 2 or just 4 We have a root in the denominator which is a no no. So we have to think of the smallest number that when multiplied by 2 will give me a perfect cube. We know that 2^3 = 8 so we can turn 2^1/3 into 8 ^ 1/3 by multiplying which gives us (2 * 4) ^ 1/3 Numerator: 3(4 ^ 1/3) Because I multiplied the denominator by x I also have to multiply the numerator by x Final Answer: [3(4^1/3)] / 4 cube root 4 can NOT cross cancel with the denominator of 4 so that is fully simplified.
I really enjoyed math throughout K-12 and into my graduate studies. I truly enjoy the reviews and the mental exercises. However, your constant talking about things that do not apply to the problem at hand runs my blood pressure up to the point that I cannot listen or watch.
But if you present your short solution to a student, it will not teach them how to arrive at your solution. John is teaching, not just showing the way an advanced level mathematician might do it in his head.
Actually 1 1/2 * ((1/2)^1/3) is not an invalid answer. Just because this guy chose to leave the radical in the denominator doesn't mean that it had to be removed in other solutions to the problem.
@@Shay-q8u Well, the task was explicitly: "simplify". 60 years ago 3*CubeRoot(4)/4 was certainly simpler to calculate than 3/(2*CubeRoot(2)), using a logarithm table -- but today, using a spreadsheet?
Thanks for discouraging me. I reamspect your opinion. But I am confused by the working that you did. Did the question ask you to simplify. You stated that your name is John but in fact your real name is MANY, since you got it wrong
It's a great question. Unfortunately, most teachers never explain why we teach certain things. The answer, BTW, to your question is a resounding 'Never'. However, this is not the point. The reason this is taught is because every time your brain is confronted with a challenge it must create new neurological connections in order to find a path to the answer. After a few years of schooling, if you allow this process to take place you will (hopefully) end up with a brain that is better able to seek out solutions.
@@mathmandrsam Well, that might be. That said, in my professional life, which includes 36 years of tax practice, and 26 years of teaching law, well, cube roots have never come up in any conversation.
@@louf7178 Yup!!! Basket Weaving, Tax, Business Law, Finance, Management, Investment Analysis, to name just a few. None of that stuff is relevant to those subject areas, and quite a few more, I would guess.
It has been over 40 years since I encountered that sort of problem Thanks for the refresher.
Well got half way through and thought I was done. I forgot about the IN in the denominators rule. So cool when I’m nudged awake again. Thankyou!❤
Split the quotient: 27^(1/3) * (1/16)^(1/3). 16 can be written as: 2 * 8. That means: 3 * (1/(2 * 8))^(1/3). 2^3 = 8. That means that the cube root of 8 is 2. So, the equation can also be written like: 3 / (2 * 2^(1/3)). Or: (3/2) / (2^(1/3)). Or: 1.5 * 2^(-1/3).
That is rounded 1.19.
That is what I got
i just came across this and it has been many years since doing this kind of math. this is what i came up with: step one- multiply by cube root of 16 / cube root of 16 to get cube root of 27 times cube root of 16 /16; step two- simplify to 3 times cube root of 8 times cube root of 2 /16; step three- simplify to 3 times 2 times cube root of 2 / 2 times 8; step four simplify to 3 times cube root of 2 / 8.
where and why did i go wrong?
The problem is in the first step. In the denominator, cube root of 16 * cube root of 16 is not 16. Almost did this myself because I am used to having a square root in the denominator, so just multiply by itself to get rid of the radical.
You all can look at it this way. When you have an a^(1/r) in the denominator, you need to give this term r-1 identical "buddies" and then put a copy of each of those "buddies" in the numerator to work toward the proper radical format when presenting a value as an answer. If it is a square root, you need one (2 - 1) "a^(1/2)" in both the numerator and denominator. if it were a seventh root then you, of course, would need six (7 - 1) "a^(1/7)"s in both the numerator and denominator.
3/4 (cube root of 4)
Thank you so much. Another excellent video! So correct about textbooks not teaching this!
I would have first simplified it as: 3*cube_root(1/16) using the multiplication property of radicals, reading 27/16 as 27 * 1/16. Then you can reduce cube_root(1/16) by using the equivalent fraction 4/64, and the division property of radicals to get 3*cube_root(4)/4
Cube root of 4/64!!
I never would have thought of that. Great stuff! Thanks.
great one. only missed the final factor. always fun though. thanks.
What are your thoughts on (3 x 2^(2/3)) / 4?
My thoughts precisely too
The way I've seen it in classes and on these math channels, is they tend to frown on having fractional exponents in the final answer, or exponents in the radicand, unless logistically necessary (ie: if it's more feasible to express the radicand as an exponentiated term rather than a rational number). Thus ∛4 is more ideal than 2^(2/3) or ∛2², but ∜71³ would be preferable to ∜357911.
I prefer the result 3/(2*cuberoot (2))
That's exactly what it should be
yep me too 16=8*2 yeilding 2*cuberoot(2)
Ah! Cannot have an irrational number in the denominator
What is the program that John uses to both have typed problems that he can write over and the hidden pen to write with? It looks like a Mac program.
With all my respect your explanation make students loose track off the main subject
3/2 x (Cube Root of 1/2)
Yes, but rationalize the denominator.
@@jamesharmon4994 Why? I have run into many equations used in engineering and science that have radicals in the denominator. If it's good enough to design bridges and spacecraft, it's good enough for any Real World application.
@@silverhammer7779 That bridge will not stand very long with a rational in the denominator.
Why would the answer be considered as simplyfied. What exactly is easier or better in the answer?
Thankyou sir ! denominator rules 🙏
It’s 3/2*cbrt(2)
That's not the final answer.
Thank you
It would make more sense to, in the last step, multiply by the cube root of 2 over the cube root of 2 twice...effectively cubing the cube root of 2 in the denominator to give 2; then simplifying the 2 cube roots of 2 in the numerator to the cube root of 4.
I got part way through doing this in my head and saw your cube root of 4 in the answer and said, "What the ...???"
I was thinking of just multiplying the cube root of 2 times the cube root of 2. But I was wrong, as you showed. Doing that will still leave a radical in the denominator. Oops!
Good problem and good explanation.
But please try not to repeat yourself so much. You explained the 7/√3 example three times.
That's where I made my mistake.
I didn't know how to rationalize a cube root denominator before watch your solution I was looking for ways to solve it this is what I thought the cube root of 16 is 16^1/3 so if I times both numerator and denominator by 16^2/3 it would leave 16^3/3 so just 16
3x 16^2/3 = 3 x cube root 256
3 x cube root of 64x4
3x4x cube of 4 / 16
3*Cube root 4 / 4
Your way is much easier
I think I did the math right, and I got (3*(4^(1/3)))/4, that is 3 times the cube root of 4, divided by 4.
After answer is shown: WOOOO!
Well explained! I got to 3/the cube root of 16 but I knew it was not enough. Repeat after me, we can't have an irrational number as the denominator. :(
Got it!
Thanks!
Takes me back to my highschool days.
I didn't get it right at first, but I understand your explanation. I am a 71-year-old senior citizen student at my local community college, with a 3.9 GPA. 😊
Thank' !
My math OCD doesn't like cubes in the numerator. I'd still like to solve it, as it seems like it doesn't have a "true" final answer. It would have a decimal that never ends, most likely. 🤔
(3cubed root of 4)/4
3/2squre root 1/2
This problem is easier if you use fractional exponents and rationalize the denominator.
Cubic of 2×3/2
Why didn't you divide the square root 4 to get 2. Thanks answer
becaue then you would have to do that to the numerator and that would cause chaos
Looks like
3/(2✔️2)
0:33
I made the same mistake. I multiplied by cube root of 2 instead of 4.
i wonder how many of his students fall asleep from boredom because he takes forever to get to the solution to the problem.
3 / (2 ⋅ ³√2)
= 3 ⋅ ²√4 / 4
why cant the answer be 3X(cube root of 2)/2 rather than 3x(cube root of 4)/4 ??
Cube root of 27/16
Cube root of 27 / cube root of 16
3 / cube root of 16
EDIT .. That wasn't it !
3(cube root of 4) / 4 is simpler ?
3/2 times ÷square rt of 2?
If your're trying to write 3/2*cuberoot(2), thats correct but you have to rationalize the denominator. To do that we multiply top and bottom by cuberoot(2)^2. cuberoots must be multiplied out 3 times unlike square roots. So apply that to the denominator, you get 2cuberoot(2) * cuberoot(2)^3 which gives 4 and on top multiply 3*cuberoot(2)^2, we square it here because there was not a cuberoot(2) already there like in the denominator. this gives 3cuberoot(4)/2
What's wrong with 3/2 × ∛(1/2)?
That will be too easy for him and not enough time wasting!
=3/2v2
its (3*cuberoot(4))/4. What you have is correct but you have to rationalize the denominator. Since we are dealing with cuberoots we have to multiply top and bottom by cuberoot(2) twice (cuberoot(2)*cuberoot(2)*cuberoot(2))=2 and on top you have 3*(cuberoot(2))^2 = (3cuberoot(4)/4)
If anyone wants to waste their time they can watch this
and anybody who wants to fail in school can follow your advice.
Dont you think that a very long route has been taken to solve the problem?
Disagree. Acceptable answer is 3/(2*cube root(2)). Reason, Sin 45 is 1/sqrt(2) not sqrt(2)/2. It is acceptable to have radicals in the denominator, well, it was when I was at secondary school 55 years ago.
For this comment, I’ll not even try to solve the question. Just a thought. ‘Many will get wrong’. In all vids. True statement, but it is phrased in the negative, missing an ‘it’ I guess on purpose, some psychology coming in… Personally, not a creator on TH-cam , I’d go with ‘Did you get it right?’. Feels much more engaging to me. Just my thoughts, let the shredding begin, as long as it stays within the boundaries of math.
1.5 all day!
so you are saying that cube of 4 is 16? not quite
I think I may have it wrong.
Many will get this wrong! How many? You must have really blessed students.
I'm astonished at the number of people watching his very basic videos. I do it as a research for viewing statistics myself.
Easy peazy.
3/(2^(1/3) 2)
I came up with 3/16.
I refuse to have anything to do with radicals.
Explanation is too much
I got the answer in about 10 seconds.
Dozed off in the middle bit. Knew the answer anyway.
It is far better to explain too much than too little.
That’s the point of the channel.
@@dave929
Congrats. I’m always on the lookout for the I did the question in nanoseconds. Here you are. Thank you for making the search easy.
√3×√3=√3²=3,is't?
The claim that Manny will get it wrong is terribly annoying. There is absolutely no need for that.
Agreed
He's right. Deal with it.
I agree!
Yeah, what did Manny ever do to become the object of cyber bullying. ✌️ Oh, the humanity.. ❤
Cubic root of 27 = 3
Cubit root of 16 = 2.52
3/2.52 = 1.19
The cube root of 16 is not 2.52. The cube root of 16 is irrational.
I'm getting (3*³√2)/4 oops should have multiplied ³√4/³√4 for my "one" used ³√2/³√2 instead (thinking squares instead of cubes. Correct answer is (3*³√4)/4.😂
Super simple I will use fraction exponents to make it easier to read.
Numerator first:
27 ^ 1/3 = (3 * 3 * 3) ^ 1/3 = 3
This shows that 27 is a perfect Cube
Denominator:
16 ^ 1/3 = (2 * 2 * 2 * 2) ^ 1/3 = 2(2^1/3)
This shows 16 is NOT a perfect cube but since 8 is we can rewrite 16^1/3 as (8 * 2) ^ 1/3. Which gives me 2(2^ 1/3)
2(2 * 4) ^ 1/3 = 2(8 ^ 1/3) = 2 * 2 or just 4
We have a root in the denominator which is a no no. So we have to think of the smallest number that when multiplied by 2 will give me a perfect cube. We know that 2^3 = 8 so we can turn 2^1/3 into 8 ^ 1/3 by multiplying which gives us (2 * 4) ^ 1/3
Numerator:
3(4 ^ 1/3)
Because I multiplied the denominator by x I also have to multiply the numerator by x
Final Answer:
[3(4^1/3)] / 4
cube root 4 can NOT cross cancel with the denominator of 4 so that is fully simplified.
1.5
I really enjoyed math throughout K-12 and into my graduate studies. I truly enjoy the reviews and the mental exercises. However, your constant talking about things that do not apply to the problem at hand runs my blood pressure up to the point that I cannot listen or watch.
3÷2√2
👍👍
That's not simplified 😄, different but at least as complicated. Honestly the starting point is cleaner
Did this in my head.
I spent few seconds to solve this simple task you speak ten minutes. Strange scool.
But if you present your short solution to a student, it will not teach them how to arrive at your solution. John is teaching, not just showing the way an advanced level mathematician might do it in his head.
Сколько лишних слов!
You do waffle on
3/(2×sqrt
These need to be 7-8 minutes in duration
1.2 if rounded to one decimal place.
Actually 1 1/2 * ((1/2)^1/3) is not an invalid answer. Just because this guy chose to leave the radical in the denominator doesn't mean that it had to be removed in other solutions to the problem.
It is standard practice to remove radicals in the denominator, and your math teacher may reject your answer, not as incorrect but as being incomplete.
try that in my class I'll sneer at you and mark the whole question wrong. You NEVER leave a root in the denominator! Even my D- students know that.
@@Shay-q8u Well, the task was explicitly: "simplify". 60 years ago 3*CubeRoot(4)/4 was certainly simpler to calculate than 3/(2*CubeRoot(2)), using a logarithm table -- but today, using a spreadsheet?
Thanks for discouraging me. I reamspect your opinion.
But I am confused by the working that you did.
Did the question ask you to simplify. You stated that your name is John but in fact your real name is MANY, since you got it wrong
Comment for the algorithm.
3/2
It might get confused many younger students when you explain too much.
This is simple:
27 is 3 cubed. And 16 is the cube of 2 :
( 3x3x3)÷ (2×2×2)
So the cube root is = 3÷2 = 1.5
16 is 2 cubed x 2. 8 is the cube of 2.
How delightful to hear from somebody without an unintelligably thick Indian accent! :P
Just like in my old s**thole of a high school back in the day...I just didn't get it. WHERE WILL I EVER USE THIS IN NON-STEM WORKING LIFE???
It's a great question. Unfortunately, most teachers never explain why we teach certain things. The answer, BTW, to your question is a resounding 'Never'. However, this is not the point. The reason this is taught is because every time your brain is confronted with a challenge it must create new neurological connections in order to find a path to the answer. After a few years of schooling, if you allow this process to take place you will (hopefully) end up with a brain that is better able to seek out solutions.
It's very basic math. But, yea, don't expect it to be used in basket weaving.
@@mathmandrsam Well, that might be. That said, in my professional life, which includes 36 years of tax practice, and 26 years of teaching law, well, cube roots have never come up in any conversation.
@@louf7178 Yup!!! Basket Weaving, Tax, Business Law, Finance, Management, Investment Analysis, to name just a few. None of that stuff is relevant to those subject areas, and quite a few more, I would guess.
You will use it when your kids ask for help with their math homework. They will think you're a genius and have more respect for you. :)
∛27 ÷ ∛16 = 3/(2∛2) · (∛4)/(∛4) = (3∛4)/4.
3/16
Answer is 3/4
You repeat the same things in all your videos. Keep it simple
3 / 2∛2
Sorry 3/2
Answer 1.5
you used a calculator and did not reduce the problem.
Waste discussion taking much time
Maths talks much less than this . Please don't bore yuor followers .
I don't think any student would want to sit and listen to you talk way too much.
If you want to skip all the BS just scroll 3/4 through the video. Or just don’t watch this channel at all.
Far too long-winded. Just get on with it already!!
John the JACKASS, knows how make a simple, 2nd or 3rd grade problem look difficult.
Too much blather and advertising.
Boring
Это ещё надо умудриться так бестолково объяснять!! Городить огород полчаса из-за плевого примера!?
Long explanation ? ? ? Be patient to understand the solution.Don’t be impatient.
Probably done that way to help keep the advertisements or commercials rolling. If click on certain spots, an advertisement automatically appears.
Answer is 1.5
You repeat the same things in all your videos. Keep it simple
Agreed but it is talking to a wall😢
3/2
3/3
You repeat the same things in all your videos. Keep it simple