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

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Math Expert
Joined: 02 Sep 2009
Posts: 55272

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06 Jul 2017, 23:30
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Difficulty:

55% (hard)

Question Stats:

59% (01:49) correct 41% (01:42) wrong based on 87 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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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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24 Apr 2019, 07:29
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Re: (2/(3a))^(1/3)   [#permalink] 24 Apr 2019, 07:29
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