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655-705 (Hard)|   Roots|      
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Bunuel
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(2/(3a))^(1/3) 06 Jul 2017, 23:30
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C
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55% (hard)

Question Stats:

60% (01:45) correct 40% (01:48) wrong based on 526 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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C
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(2/(3a))^(1/3) 06 Jul 2017, 23:54
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Substitute A=2 in Question stem

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

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pushpitkc
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(2/(3a))^(1/3) 07 Jul 2017, 00:04
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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) 07 Jul 2017, 05:59
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Bunuel
\(\sqrt[3]{\frac{2}{3a}}\)


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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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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(2/(3a))^(1/3) 12 Jul 2017, 16:21
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Bunuel
\(\sqrt[3]{\frac{2}{3a}}\)


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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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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(2/(3a))^(1/3) 02 Jul 2019, 08:23
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Bunuel
\(\sqrt[3]{\frac{2}{3a}}\)


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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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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(2/(3a))^(1/3) 05 Jul 2019, 13:13
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Bunuel
\(\sqrt[3]{\frac{2}{3a}}\)


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

Leaving our answer as 2.
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(2/(3a))^(1/3) 05 Jul 2019, 14:54
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omavsp
Bunuel
\(\sqrt[3]{\frac{2}{3a}}\)


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!
Show more
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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(2/(3a))^(1/3) 14 Aug 2019, 04:30
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Bunuel
\(\sqrt[3]{\frac{2}{3a}}\)

A. \(\frac{\sqrt[3]{6a}}{3a}\)
B. \(\frac{\sqrt[3]{18a}}{3a}\)
C. \(\frac{\sqrt[3]{18a^2}}{3a}\)
D. 2
E. 6a
Show more

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

Answer (C).
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(2/(3a))^(1/3) 09 Sep 2024, 21:44
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