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

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(2/(3a))^(1/3)  [#permalink]

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New post 06 Jul 2017, 23:30
00:00
A
B
C
D
E

Difficulty:

  55% (hard)

Question Stats:

60% (01:43) correct 40% (01:36) wrong based on 73 sessions

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Re: (2/(3a))^(1/3)  [#permalink]

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New post 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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Re: (2/(3a))^(1/3)  [#permalink]

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New post 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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(2/(3a))^(1/3)  [#permalink]

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New post 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;

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

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

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

Answer (C)...
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Re: (2/(3a))^(1/3)  [#permalink]

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New post 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

Answer: C
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Re: (2/(3a))^(1/3)  [#permalink]

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New post 24 Apr 2019, 07:29
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Re: (2/(3a))^(1/3)   [#permalink] 24 Apr 2019, 07:29
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