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[m]\frac{1}{\frac{1}{\sqrt[3]{a+b}}}*\frac{\sqrt[3]{(a+b})^4}{\frac{1}

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[m]\frac{1}{\frac{1}{\sqrt[3]{a+b}}}*\frac{\sqrt[3]{(a+b})^4}{\frac{1}  [#permalink]

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New post Updated on: 11 Nov 2014, 09:43
2
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A
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D
E

Difficulty:

  35% (medium)

Question Stats:

70% (01:34) correct 30% (02:12) wrong based on 110 sessions

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\(\frac{1}{\frac{1}{\sqrt[3]{a+b}}}*\frac{\sqrt[3]{(a+b)^4}}{\frac{1}{(a+b)^{-\frac{2}{3}}}}=\)

(A) 1
(B) (a + b)-1/3
(C) (a + b)1/3
(D) (a + b)2/3
(E) a + b

Sorry about the image equation. Tried hard with latex to make this appear fine, but doesn't seem to be working out

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Originally posted by rajarams on 11 Nov 2014, 09:15.
Last edited by Bunuel on 11 Nov 2014, 09:43, edited 2 times in total.
Edited the question.
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[m]\frac{1}{\frac{1}{\sqrt[3]{a+b}}}*\frac{\sqrt[3]{(a+b})^4}{\frac{1}  [#permalink]

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New post 11 Nov 2014, 09:33
3
1
rajarams wrote:
Evaluate the expression:

Image

(A) 1
(B) (a + b)-1/3
(C) (a + b)1/3
(D) (a + b)2/3
(E) a + b

Sorry about the image equation. Tried hard with latex to make this appear fine, but doesn't seem to be working out


\(\frac{1}{\frac{1}{\sqrt[3]{a+b}}}*\frac{\sqrt[3]{(a+b)^4}}{\frac{1}{(a+b)^{-\frac{2}{3}}}}=\)


\(\sqrt[3]{a+b}*\sqrt[3]{(a+b)^4}*(a+b)^{-\frac{2}{3}}=\)


\(\sqrt[3]{(a+b)(a+b)^4}*\frac{1}{(a+b)^{\frac{2}{3}}}=\)


\(\sqrt[3]{(a+b)^5}*\frac{1}{\sqrt[3]{(a+b)^2}}=\)


\(\sqrt[3]{\frac{(a+b)^5}{(a+b)^2}=\)


\(\sqrt[3]{(a+b)^3}=a+b\)

Answer: E.

Hope it's clear.
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Re: [m]\frac{1}{\frac{1}{\sqrt[3]{a+b}}}*\frac{\sqrt[3]{(a+b})^4}{\frac{1}  [#permalink]

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New post 11 Nov 2014, 09:54
Bunuel wrote:
rajarams wrote:
Evaluate the expression:

Image

(A) 1
(B) (a + b)-1/3
(C) (a + b)1/3
(D) (a + b)2/3
(E) a + b

Sorry about the image equation. Tried hard with latex to make this appear fine, but doesn't seem to be working out


\(\frac{1}{\frac{1}{\sqrt[3]{a+b}}}*\frac{\sqrt[3]{(a+b)^4}}{\frac{1}{(a+b)^{-\frac{2}{3}}}}=\)


\(\sqrt[3]{a+b}*\sqrt[3]{(a+b)^4}*(a+b)^{-\frac{2}{3}}=\)


\(\sqrt[3]{(a+b)(a+b)^4}*\frac{1}{(a+b)^{\frac{2}{3}}}=\)


\(\sqrt[3]{(a+b)^5}*\frac{1}{\sqrt[3]{(a+b)^2}}=\)


\(\sqrt[3]{\frac{(a+b)^5}{(a+b)^2}=\)


\(\sqrt[3]{(a+b)^3}=a+b\)

Answer: E.

Hope it's clear.


Thanks Bunuel, could you put an explanation of what you are doing for each step as you reduce the fractions?
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Re: [m]\frac{1}{\frac{1}{\sqrt[3]{a+b}}}*\frac{\sqrt[3]{(a+b})^4}{\frac{1}  [#permalink]

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New post 11 Nov 2014, 20:49
1
Answer = E

\(\frac{1}{\frac{1}{\sqrt[3]{a+b}}}*\frac{\sqrt[3]{(a+b)^4}}{\frac{1}{(a+b)^{-\frac{2}{3}}}}=\)

Get rid of the \(\sqrt{SQUARE ROOT}\) sign first

\(\frac{1}{\frac{1}{{(a+b)}^{\frac{1}{3}}}}*\frac{{(a+b)^{\frac{4}{3}}}}{\frac{1}{(a+b)^{-\frac{2}{3}}}}=\)

Normalize the denominator (Adjust power sign)

\(\frac{1}{{(a+b)}^{\frac{-1}{3}}}*\frac{{(a+b)^{\frac{4}{3}}}}{{(a+b)^{\frac{2}{3}}}=\)

Get rid of the denominator

\({{(a+b)}^{\frac{+1}{3}}} * (a+b)^{\frac{4}{3}} * (a+b)^{\frac{-2}{3}} =\)

Bases are the same; add the powers

\({{(a+b)}^{(\frac{+1}{3} - \frac{2}{3} + \frac{4}{3})} =\)

a+b
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Re: [m]\frac{1}{\frac{1}{\sqrt[3]{a+b}}}*\frac{\sqrt[3]{(a+b})^4}{\frac{1}  [#permalink]

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Re: [m]\frac{1}{\frac{1}{\sqrt[3]{a+b}}}*\frac{\sqrt[3]{(a+b})^4}{\frac{1}   [#permalink] 10 Dec 2018, 11:39
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