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

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

Kudos please! Mr Bunuel, the quant king!!

D is not correct. Anyone else wants to try?

for a=64
the sol for b = +/- 8. This gives two values of B, therefore insufficient.

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)

——-
Shouldn’t B also be sufficient?

So I solved it using a^2-b^2 formula and got:
(4+b^1/3)(4-b^1/3) = 12
(Cube root that is)

b^1/3 could only be 2 right and not -2 cuz we say that the solution of a root can only be positive - shouldn’t the same apply to this too?

Or is it because it’s an odd power of root (ie cube root), it is possible to have -2 also? Pls let me know if I’m not thinking on correct lines Bunuel

Posted from my mobile device

Let us consider \(\sqrt[3]{a} = x\) and \(\sqrt[3]{b} = y\).

Hence, the given statement \(\sqrt[3]{a}^2 - \sqrt[3]{b}^2 =12\) now becomes \(x^2 - y^2 =12 \)=> \((x+y)(x-y)=12\)--> Equation 1
and the question we have to answer is \(x+ y =?\) --> Equation 2

Now these equations look like something we can work with.

Statement 1:

\(\sqrt[3]{a} = \sqrt[3]{b}+ 2\) => \(x = y+2\) --> Equation 3
Let us use equation 3 in 1. Doing so gives us:-
\((x+y)(x-y)=12\)
\((y+2+y)(y+2-y)=12\)
\((2y+2)(2)=12\)
\(2y+2=6\)
y=2
Let us use value of y in Equation 3 to get value of x.
\(x=y+2\)
\(x=4.\)

Hence, using x and y in Equation 2 we can get a definite answer. Statement 1 is sufficient.

Statement 2:
\(a=64\)
Converting a in terms of x, we get
=> \(x^3=64 \)=> \(x=4\)

Using the x=4 in Equation 1 gives us:-
\(4^2-y^2=12\)
\(4=y^2\)
\(y=2\) or \(y=-2\)

If \(x=4\) and \(y=2\), Equation 2 gives us value of 6.
If \(x=4\) and \(y=-2\), Equation 2 gives us value of 2.
Since we do not have a definite answer, Statement 2 is not sufficient.

Hence, answer is A.

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