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If a^{2/3} - b^{2/3} = 12, then \sqrt[3]{a} + \sqrt[3]{b} = ?

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If a^{2/3} - b^{2/3} = 12, then \sqrt[3]{a} + \sqrt[3]{b} = ? [#permalink]

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New post 06 Nov 2014, 09:02
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Tough and Tricky questions: Algebra.



If \(a^{\frac{2}{3}} - b^{\frac{2}{3}} = 12\), then \(\sqrt[3]{a} + \sqrt[3]{b} =\) ?

(1) \(\sqrt[3]{a} = \sqrt[3]{b} + 2\)

(2) a = 64

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[Reveal] Spoiler: OA

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Re: If a^{2/3} - b^{2/3} = 12, then \sqrt[3]{a} + \sqrt[3]{b} = ? [#permalink]

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New post 06 Nov 2014, 13:39
Both are sufficient

statement 1: use a^2-b^2=(a+b)(a-b). Tis gives us answer 6.
Sufficient

Statement 2
Put 64 in first equation. that gives us b=8.
Put both a and b into final question and find answer as 6.
Sufficient
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Re: If a^{2/3} - b^{2/3} = 12, then \sqrt[3]{a} + \sqrt[3]{b} = ? [#permalink]

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Given:
\(a^{2/3}-b^{2/3}= (a^{1/3}+b^{1/3})*(a^{1/3}-b^{1/3}) = 12\)

\(a^{1/3}+b^{1/3} = ?\)

Statement 1:

\((a^{1/3}-b^{1/3})=2\)

Therefore \(a^{1/3}+b^{1/3} = 6\)

Sufficient.

Statement 2:

a=64

This can be used to solve for b and hence is sufficient.

Answer : D

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Re: If a^{2/3} - b^{2/3} = 12, then \sqrt[3]{a} + \sqrt[3]{b} = ? [#permalink]

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New post 10 Nov 2014, 05:35
Expert's post
D is not correct. Anyone else wants to try?
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Re: If a^{2/3} - b^{2/3} = 12, then \sqrt[3]{a} + \sqrt[3]{b} = ? [#permalink]

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New post 10 Nov 2014, 05:55
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As per question : a^(2/3)-b^(2/3)=12........(1)

so we can break this in terms of a^2-b^2=(a+b)(a-b)
(a^1/3)^2-(b^1/3)^2 = (a^1/3 +b^1/3)(a^1/3 - b^1/3)=12 ......(2)

1) a^1/3 = b^1/3 +2
a^1/3 - b^1/3 =2
So inserting this value in equation (2) will solve this puzzle. SUFFICIENT

2) a=64
taking this value in equation (1)
64^2/3 - b^2/3=12
16 -12 =b^2/3
so b^2/3=4 and b^1/3=+-2.
this two different values for b^1/3 will create new puzzle rather solving it.INSUFFICIENT

Correct answer :A
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Re: If a^{2/3} - b^{2/3} = 12, then \sqrt[3]{a} + \sqrt[3]{b} = ? [#permalink]

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New post 14 Apr 2015, 04:16
for a=64
the sol for b = +/- 8. This gives two values of B, therefore insufficient.
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Re: If a^{2/3} - b^{2/3} = 12, then \sqrt[3]{a} + \sqrt[3]{b} = ? [#permalink]

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New post 08 Nov 2017, 14:00
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Answer (A)

\(a ^ {2/3} - b ^ {2/3} = 12\)

Let \(x = a ^ {1/3}, y = b ^ {1/3}\) => \(x ^ 2 - y ^ 2 = 12\) =>\((x + y)(x - y)\) = 12

Question :\(a ^ {1/3} + b ^ {1/3}\) = ? => \(x + y =\)?

Let us attack statements

Statement 1:

\(a ^ {1/3} = b ^ {1/3} + 2\) => \(x = y + 2\) => \(x - y = 2\)

we have \((x + y) (x - y) = 12\)
so \(x + y = 6\) -> sufficient

Statement 2:
\(a = 64\) => taking cube roots on both sides => \(a ^ {1/3}\) = 4 => \(x = 4\)

using, \(x ^ 2 - y ^ 2 = 12\) => \(4 ^ 2 - y ^ 2 = 12\)=> \(y ^ 2 = 4\)=>\(y = 2 or -2\)

so \(x + y = 4+2 = 6\) or \(x - y = 4-2 = 2\) => Not sufficient

Answer (A)
Re: If a^{2/3} - b^{2/3} = 12, then \sqrt[3]{a} + \sqrt[3]{b} = ?   [#permalink] 08 Nov 2017, 14:00
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