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(2/(3a))^(1/3)

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Math Expert
Joined: 02 Sep 2009
Posts: 59588

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06 Jul 2017, 23:30
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Question Stats:

60% (01:45) correct 40% (01:45) wrong based on 159 sessions

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$$\sqrt[3]{\frac{2}{3a}}$$

A. $$\frac{\sqrt[3]{6a}}{3a}$$

B. $$\frac{\sqrt[3]{18a}}{3a}$$

C. $$\frac{\sqrt[3]{18a^2}}{3a}$$

D. 2

E. 6a

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Joined: 07 Mar 2019
Posts: 28

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05 Jul 2019, 14:54
2
omavsp wrote:
Bunuel wrote:
$$\sqrt[3]{\frac{2}{3a}}$$

A. $$\frac{\sqrt[3]{6a}}{3a}$$

B. $$\frac{\sqrt[3]{18a}}{3a}$$

C. $$\frac{\sqrt[3]{18a^2}}{3a}$$

D. 2

E. 6a

Dear Bunuel,

can you please explain how people here came up with Multiplying and dividing by $$(3a)^2$$.

Im having hard time with reaching the right numbers in such circumstances.

Thank you so much!

Saying you multiply the numerator and denominator by (3a)^2 is a poor explanation, they meant to say to multiply it by ((3a)^2)^1/3.

Taking the cubed root is the equivalent of raising x^1/3. Thus, to make the denominator 3a, we need to multiply it by (3a)^2/3, since 1/3 + 2/3 = 1, in which case it would be (3a)^1.

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Joined: 06 May 2015
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06 Jul 2017, 23:54
Substitute A=2 in Question stem

Then select the answer which gives same value for a=2

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07 Jul 2017, 00:04
1
We are given the expression $$\sqrt[3]{\frac{2}{3a}}$$

Multiplying and dividing by $$(3a)^2$$

we will get $$\frac{\sqrt[3]{18a^2}}{\sqrt[3]{27a^3}}$$ = $$\frac{\sqrt[3]{18a^2}}{\sqrt[3]{(3a)^3}}$$ = $$\frac{\sqrt[3]{18a^2}}{3a}$$(Option C)
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07 Jul 2017, 05:59
1
Bunuel wrote:
$$\sqrt[3]{\frac{2}{3a}}$$

A. $$\frac{\sqrt[3]{6a}}{3a}$$

B. $$\frac{\sqrt[3]{18a}}{3a}$$

C. $$\frac{\sqrt[3]{18a^2}}{3a}$$

D. 2

E. 6a

Given the expression $$\sqrt[3]{\frac{2}{3a}}$$

Multiplying numerator and denominator with $$(3a)^2$$, we get;

$$\sqrt[3]{\frac{2*(3a)^2}{3a*(3a)^2}}$$

$$\sqrt[3]{\frac{2*9a^2}{(3a)^3}}$$

$$\frac{\sqrt[3]{18a^2}}{3a}$$

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12 Jul 2017, 16:21
Bunuel wrote:
$$\sqrt[3]{\frac{2}{3a}}$$

A. $$\frac{\sqrt[3]{6a}}{3a}$$

B. $$\frac{\sqrt[3]{18a}}{3a}$$

C. $$\frac{\sqrt[3]{18a^2}}{3a}$$

D. 2

E. 6a

In order to rationalize the denominator of this cube root problem, we must have a perfect cube in the denominator of the fraction. Thus, we have to multiply the numerator and denominator by the square of the denominator:

3^√(2/3a) * 3^√(9a^2/9a^2)

3^√(18a^2/27a^3)

3^√(18a^2)/3^√(27a^3)

3^√(18a^2)/3a

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Joined: 20 Aug 2017
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02 Jul 2019, 08:23
Bunuel wrote:
$$\sqrt[3]{\frac{2}{3a}}$$

A. $$\frac{\sqrt[3]{6a}}{3a}$$

B. $$\frac{\sqrt[3]{18a}}{3a}$$

C. $$\frac{\sqrt[3]{18a^2}}{3a}$$

D. 2

E. 6a

Dear Bunuel,

can you please explain how people here came up with Multiplying and dividing by $$(3a)^2$$.

Im having hard time with reaching the right numbers in such circumstances.

Thank you so much!
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Joined: 07 May 2018
Posts: 61

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05 Jul 2019, 13:13
JeffTargetTestPrep wrote:
Bunuel wrote:
$$\sqrt[3]{\frac{2}{3a}}$$

A. $$\frac{\sqrt[3]{6a}}{3a}$$

B. $$\frac{\sqrt[3]{18a}}{3a}$$

C. $$\frac{\sqrt[3]{18a^2}}{3a}$$

D. 2

E. 6a

In order to rationalize the denominator of this cube root problem, we must have a perfect cube in the denominator of the fraction. Thus, we have to multiply the numerator and denominator by the square of the denominator:

3^√(2/3a) * 3^√(9a^2/9a^2)

3^√(18a^2/27a^3)

3^√(18a^2)/3^√(27a^3)

3^√(18a^2)/3a

Hi why can't we not raise this function to the power of 3: ((2/3a)^1/3)^3

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14 Aug 2019, 04:30
Bunuel wrote:
$$\sqrt[3]{\frac{2}{3a}}$$

A. $$\frac{\sqrt[3]{6a}}{3a}$$
B. $$\frac{\sqrt[3]{18a}}{3a}$$
C. $$\frac{\sqrt[3]{18a^2}}{3a}$$
D. 2
E. 6a

rationalize the denominator:
$$\sqrt[3]{\frac{2}{3a}}•\sqrt[3]{\frac{(3a)^2}{(3a)^2}}=\frac{\sqrt[3]{2(3a)^2}}{3a}=\frac{\sqrt[3]{18a^2}}{3a}$$

Re: (2/(3a))^(1/3)   [#permalink] 14 Aug 2019, 04:30
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