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

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Math Expert
Joined: 02 Sep 2009
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12 May 2017, 04:49
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61% (01:01) correct 39% (00:57) wrong based on 182 sessions

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$$\frac{5x^2}{\sqrt[3]{x^2}$$

A. $$5\sqrt[3]{x}$$

B. $$5x\sqrt[3]{x}$$

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

D. $$5x^2\sqrt{x^2}$$

E. $$125x^4$$

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12 May 2017, 04:59
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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. $$125x^4$$

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

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13 May 2017, 03:10
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. $$125x^4$$

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

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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13 May 2017, 03:28
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. $$125x^4$$

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

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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13 May 2017, 03:33
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. $$125x^4$$

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

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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Joined: 11 Feb 2017
Posts: 202

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13 May 2017, 03:50
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. $$125x^4$$

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

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]

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13 May 2017, 04:33
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..

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

In case that you want to get rid of cube root of $$\sqrt[3]{x}$$
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13 May 2017, 12:46
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..

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

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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13 May 2017, 19:17
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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14 May 2017, 00:45
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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14 May 2017, 01:49
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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14 May 2017, 02:10
[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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14 May 2017, 02:20
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?

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.
$$\sqrt[3]{8x^3}=1 \implies 8x^3=1$$

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25 Apr 2018, 11:43
@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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25 Apr 2018, 13:40
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. $$125x^4$$

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

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.
Re: 5x^2/(x^2)^(1/3)   [#permalink] 25 Apr 2018, 13:40
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