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

Kudos for a correct solution.
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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

D is not correct. Anyone else wants to try?
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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What are GMAT Club Tests?
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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)