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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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Manager
Joined: 31 Oct 2011
Posts: 50
Location: India
GMAT 1: 660 Q48 V32
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Updated on: 11 Nov 2014, 09:43
2
00:00

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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Joined: 02 Sep 2009
Posts: 62351

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

(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

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

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

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

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

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

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

Hope it's clear.
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11 Nov 2014, 09:54
Bunuel wrote:
rajarams wrote:
Evaluate the expression:

(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

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

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

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

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

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

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

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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11 Nov 2014, 20:49
1

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

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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10 Dec 2018, 11:39
Hello from the GMAT Club BumpBot!

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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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