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5x^2/(x^2)^(1/3)

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5x^2/(x^2)^(1/3)  [#permalink]

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New post 12 May 2017, 03:49
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Re: 5x^2/(x^2)^(1/3)  [#permalink]

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New post 12 May 2017, 03:59
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Bunuel wrote:
\(\frac{5x^2}{\sqrt[3]{x^2}\)


A. \(5\sqrt[3]{x}\)

B. \(5x\sqrt[3]{x}\)

C. \(5x\sqrt[3]{x^2}\)

D. \(5x^2\sqrt{x^2}\)

E. \(125x^4\)



\(\frac{5x^2}{\sqrt[3]{x^2}\)
= \(\frac{5x^2}{x^{2/3}}\)
= \(5 x^{2-2/3}\)
= \(5 x^{4/3}\)
= \(5x^{1+1/3}\)
= \(5x\sqrt[3]{x}\)

Answer B
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Re: 5x^2/(x^2)^(1/3)  [#permalink]

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New post 13 May 2017, 02:10
shashankism wrote:
Bunuel wrote:
\(\frac{5x^2}{\sqrt[3]{x^2}\)


A. \(5\sqrt[3]{x}\)

B. \(5x\sqrt[3]{x}\)

C. \(5x\sqrt[3]{x^2}\)

D. \(5x^2\sqrt{x^2}\)

E. \(125x^4\)



\(\frac{5x^2}{\sqrt[3]{x^2}\)
= \(\frac{5x^2}{x^{2/3}}\)
= \(5 x^{2-2/3}\)
= \(5 x^{4/3}\)
= \(5x^{1+1/3}\)
= \(5x\sqrt[3]{x}\)

Answer B





What is wrong with cubing numerator and denominator?

It will be
125 * (x)power 6 / (x)power 2
Ans will be

125 * (x)power 4
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Re: 5x^2/(x^2)^(1/3)  [#permalink]

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New post 13 May 2017, 02:28
rocko911 wrote:
shashankism wrote:
Bunuel wrote:
\(\frac{5x^2}{\sqrt[3]{x^2}\)


A. \(5\sqrt[3]{x}\)

B. \(5x\sqrt[3]{x}\)

C. \(5x\sqrt[3]{x^2}\)

D. \(5x^2\sqrt{x^2}\)

E. \(125x^4\)



\(\frac{5x^2}{\sqrt[3]{x^2}\)
= \(\frac{5x^2}{x^{2/3}}\)
= \(5 x^{2-2/3}\)
= \(5 x^{4/3}\)
= \(5x^{1+1/3}\)
= \(5x\sqrt[3]{x}\)

Answer B





What is wrong with cubing numerator and denominator?

It will be
125 * (x)power 6 / (x)power 2
Ans will be

125 * (x)power 4


Hey Rocko911,

Why do you want to take cube of the given expression.. Its no where written to find cube.
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Re: 5x^2/(x^2)^(1/3)  [#permalink]

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New post 13 May 2017, 02:33
1
rocko911 wrote:
shashankism wrote:
Bunuel wrote:
\(\frac{5x^2}{\sqrt[3]{x^2}\)


A. \(5\sqrt[3]{x}\)

B. \(5x\sqrt[3]{x}\)

C. \(5x\sqrt[3]{x^2}\)

D. \(5x^2\sqrt{x^2}\)

E. \(125x^4\)



\(\frac{5x^2}{\sqrt[3]{x^2}\)
= \(\frac{5x^2}{x^{2/3}}\)
= \(5 x^{2-2/3}\)
= \(5 x^{4/3}\)
= \(5x^{1+1/3}\)
= \(5x\sqrt[3]{x}\)

Answer B





What is wrong with cubing numerator and denominator?

It will be
125 * (x)power 6 / (x)power 2
Ans will be

125 * (x)power 4




If you want to solve by cubing the expression then you will have to take cube root of the same..

\(\frac{5x^2}{\sqrt[3]{x^2}\)
=\(\sqrt[3]{(\frac{5x^2}{\sqrt[3]{x^2})^3}\)
=\(\sqrt[3]{(\frac{125x^6}{x^2})}\)
=\(\sqrt[3]{125x^4}\)
= \(5x\sqrt[3]{x}\)
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Re: 5x^2/(x^2)^(1/3)  [#permalink]

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New post 13 May 2017, 02:50
shashankism wrote:
shashankism wrote:
Bunuel wrote:
\(\frac{5x^2}{\sqrt[3]{x^2}\)


A. \(5\sqrt[3]{x}\)

B. \(5x\sqrt[3]{x}\)

C. \(5x\sqrt[3]{x^2}\)

D. \(5x^2\sqrt{x^2}\)

E. \(125x^4\)



\(\frac{5x^2}{\sqrt[3]{x^2}\)
= \(\frac{5x^2}{x^{2/3}}\)
= \(5 x^{2-2/3}\)
= \(5 x^{4/3}\)
= \(5x^{1+1/3}\)
= \(5x\sqrt[3]{x}\)

Answer B





What is wrong with cubing numerator and denominator?

It will be
125 * (x)power 6 / (x)power 2
Ans will be

125 * (x)power 4




If you want to solve by cubing the expression then you will have to take cube root of the same..

\(\frac{5x^2}{\sqrt[3]{x^2}\)
=\(\sqrt[3]{(\frac{5x^2}{\sqrt[3]{x^2})^3}\)
=\(\sqrt[3]{(\frac{125x^6}{x^2})}\)
=\(\sqrt[3]{125x^4}\)
= \(5x\sqrt[3]{x}\)[/quote]



Why do we need to take cube root???? Please help
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Re: 5x^2/(x^2)^(1/3)  [#permalink]

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New post 13 May 2017, 03:33
rocko911 wrote:
shashankism wrote:
rocko911 wrote:
What is wrong with cubing numerator and denominator?
It will be
125 * (x)power 6 / (x)power 2
Ans will be

125 * (x)power 4

If you want to solve by cubing the expression then you will have to take cube root of the same..

\(\frac{5x^2}{\sqrt[3]{x^2}}\)
=\(\sqrt[3]{\Bigg (\frac{5x^2}{\sqrt[3]{x^2}}\Bigg )^3}\)
=\(\sqrt[3]{(\frac{125x^6}{x^2})}\)
=\(\sqrt[3]{125x^4}\)
= \(5x\sqrt[3]{x}\)


Why do we need to take cube root???? Please help


In case that you want to get rid of cube root of \(\sqrt[3]{x}\)
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Re: 5x^2/(x^2)^(1/3)  [#permalink]

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New post 13 May 2017, 11:46
nguyendinhtuong wrote:
rocko911 wrote:
shashankism wrote:
What is wrong with cubing numerator and denominator?
It will be
125 * (x)power 6 / (x)power 2
Ans will be

125 * (x)power 4

If you want to solve by cubing the expression then you will have to take cube root of the same..

\(\frac{5x^2}{\sqrt[3]{x^2}}\)
=\(\sqrt[3]{\Bigg (\frac{5x^2}{\sqrt[3]{x^2}}\Bigg )^3}\)
=\(\sqrt[3]{(\frac{125x^6}{x^2})}\)
=\(\sqrt[3]{125x^4}\)
= \(5x\sqrt[3]{x}\)


Why do we need to take cube root???? Please help


In case that you want to get rid of cube root of \(\sqrt[3]{x}\)[/quote]



yeah to get rid of \(\sqrt[3]{x}\) we are cubing the whole expression.... but as told by Shashankism, He cubed the whole problem AND took CUBE ROOT ALSO.......SO WHY BOTH? Why not only cubing the whole expression and get rid of \(\sqrt[3]{x}\) ??????????
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Re: 5x^2/(x^2)^(1/3)  [#permalink]

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New post 13 May 2017, 18:17
rocko911 wrote:
yeah to get rid of \(\sqrt[3]{x}\) we are cubing the whole expression.... but as told by Shashankism, He cubed the whole problem AND took CUBE ROOT ALSO.......SO WHY BOTH? Why not only cubing the whole expression and get rid of \(\sqrt[3]{x}\) ??????????


He did so to make the expression more simple for calculate. You could make cube only of \(\sqrt[3]{x^2}\), however, you still have to go around to come to the answer.
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New post 13 May 2017, 23:45
nguyendinhtuong wrote:
rocko911 wrote:
yeah to get rid of \(\sqrt[3]{x}\) we are cubing the whole expression.... but as told by Shashankism, He cubed the whole problem AND took CUBE ROOT ALSO.......SO WHY BOTH? Why not only cubing the whole expression and get rid of \(\sqrt[3]{x}\) ??????????


He did so to make the expression more simple for calculate. You could make cube only of \(\sqrt[3]{x^2}\), however, you still have to go around to come to the answer.





No, i dont get it

See how I did this..

5x^2 / (x^2)^(1/3)

5x(x^2) / x^(2/3)

CUBING NUMERATOR AND DENOMINATOR

125 x^(6) / x^(2)

125 x^4

what is wrong with this ?
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Re: 5x^2/(x^2)^(1/3)  [#permalink]

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New post 14 May 2017, 00:49
rocko911 wrote:
nguyendinhtuong wrote:
rocko911 wrote:
yeah to get rid of \(\sqrt[3]{x}\) we are cubing the whole expression.... but as told by Shashankism, He cubed the whole problem AND took CUBE ROOT ALSO.......SO WHY BOTH? Why not only cubing the whole expression and get rid of \(\sqrt[3]{x}\) ??????????


He did so to make the expression more simple for calculate. You could make cube only of \(\sqrt[3]{x^2}\), however, you still have to go around to come to the answer.


No, i dont get it

See how I did this..

5x^2 / (x^2)^(1/3) \(\quad =\frac{5x^2}{(x^2)^\frac{1}{3}}\) OK

5x(x^2) / x^(2/3) \(\quad =\frac{5 \times x \times x^2}{x^\frac{2}{3}}\) Wrong. You mean \(=\frac{5 \times x^2}{x^\frac{2}{3}}\) ?

CUBING NUMERATOR AND DENOMINATOR

125 x^(6) / x^(2) .....Hey, this step is wrong. You can't left the cube root sign like this. You have to keep it like this:
\(=\sqrt[3]{\Bigg ( \frac{5 \times x^2}{x^\frac{2}{3}} \Bigg ) ^3}=\sqrt[3]{\frac{125 \times x^6}{x^2}}\)


125 x^4 .....As this step led to incorrect value. This step must be \(=\sqrt[3]{125x^4}\)

what is wrong with this ?


Your way is wrong as I explained above
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Re: 5x^2/(x^2)^(1/3)  [#permalink]

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New post 14 May 2017, 01:10
[quote="nguyendinhtuong"][quote="rocko911"][quote="nguyendinhtuong"][quote="rocko911"]



5×x2x23=5×x2x23 ?

CUBING NUMERATOR AND DENOMINATOR

125 x^(6) / x^(2) .....Hey, this step is wrong. You can't left the cube root sign like this. You have to keep it like this:
=(5×x2x23)3−−−−−−−−⎷3=125×x6x2−−−−−−√3=(5×x2x23)33=125×x6x23



125 x^4 .....As this step led to incorrect value. This step must be =125x4−−−−−√3



cube root was in denominator and when i already did x^(2/3) from where did the cube and a cube root came from?????????? i dont get it,, can u explain by basic and step by step approach?
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Re: 5x^2/(x^2)^(1/3)  [#permalink]

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New post 14 May 2017, 01:20
rocko911 wrote:
cube root was in denominator and when i already did x^(2/3) from where did the cube and a cube root came from?????????? i dont get it,, can u explain by basic and step by step approach?


I don't get your question.

Let make another example. For example, simplify this \(\sqrt[3]{8x^3}\)

The correct way is \(\sqrt[3]{8x^3}=\sqrt[3]{(2x)^3}=2x\)

Your way is like this \(\sqrt[3]{8x^3}=8x^3\). You left out cube root and get wrong value.

However, you could left out cube when solving equation.
\(\sqrt[3]{8x^3}=1 \implies 8x^3=1\)

Hope this clears your problem
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Re: 5x^2/(x^2)^(1/3)  [#permalink]

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New post 25 Apr 2018, 10:43
@Bunel can someone explain why the answer is not C? I dont fully understand how the math was done to get B.
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Re: 5x^2/(x^2)^(1/3)  [#permalink]

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New post 25 Apr 2018, 12:40
rocko911 wrote:
shashankism wrote:
Bunuel wrote:
\(\frac{5x^2}{\sqrt[3]{x^2}\)


A. \(5\sqrt[3]{x}\)

B. \(5x\sqrt[3]{x}\)

C. \(5x\sqrt[3]{x^2}\)

D. \(5x^2\sqrt{x^2}\)

E. \(125x^4\)



\(\frac{5x^2}{\sqrt[3]{x^2}\)
= \(\frac{5x^2}{x^{2/3}}\)
= \(5 x^{2-2/3}\)
= \(5 x^{4/3}\)
= \(5x^{1+1/3}\)
= \(5x\sqrt[3]{x}\)

Answer B





What is wrong with cubing numerator and denominator?

It will be
125 * (x)power 6 / (x)power 2
Ans will be

125 * (x)power 4


In this case you are not necessarily multiplying the expression by 1, so you have created a different term by cubing both the numerator and denominator and then simplifying. So taking the cube root of your solution 125x^4 should get you back to to the answer.
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Re: 5x^2/(x^2)^(1/3)  [#permalink]

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Re: 5x^2/(x^2)^(1/3)   [#permalink] 10 Jan 2020, 04:42
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