Hi experts
Bunuel KarishmaB, I was wondering if you could please help me point what I am doing wrong here. To evaluate statement 2, I would actually do the following to the stimulus:
Cube root all terms and numbers in the numerator and denominator so that the expression turns into the following:
Cube root(x^3) + Cube root(y^3) +Cube root(z^3) / Cube root(xyz)
*I did this because even though I can't take the square root of a variable with an even exponent because I don't know if the result will be positive or negative. Since we are working here with an odd exponent, i figured that there will only be 1 case, either the result is positive or negative.
After simplifying, the expression would turn into:
x+y+z/cube root(xyz).
Since the second statement tells me that x+y+z=0, then the result will be:
0/cube root(xyz)
Thus, the result of the expression will always be 0.
Many thanks in advance for your help.
Bunuel
9. If \(x\), \(y\) and \(z\) are non-zero numbers, what is the value of \(\frac{x^3+y^3+z^3}{xyz}\)?
(1) \(xyz=-6\)
(2) \(x+y+z=0\)
(1) \(xyz=-6\) --> infinitely many combinations of \(x\), \(y\) and \(z\) are possible which will give different values of the expression in the stem: try x=y=1 and y=-6 or x=1, y=2, z=-3. Not sufficient.
(2) \(x+y+z=0\) --> \(x=-(y+z)\) --> substitute this value of x into the expression in the stem --> \(\frac{x^3+y^3+z^3}{xyz}=\frac{-(y+z)^3+y^3+z^3}{xyz}=\frac{-y^3-3y^2z-3yz^2-z^3+y^3+z^3}{xyz}=\frac{-3y^2z-3yz^2}{xyz}=\frac{-3yz(y+z)}{xyz}\), as \(x=-(y+z)\) then: \(\frac{-3yz(y+z)}{xyz}=\frac{-3yz*(-x)}{xyz}=\frac{3xyz}{xyz}=3\). Sufficient.
Must know for the GMAT: \((x+y)^3=(x+y)(x^2+2xy+y^2)=x^3+3x^2y+3xy^2+y^3\) and \((x-y)^3=(x-y)(x^2-2xy+y^2)=x^3-3x^2y+3xy^2-y^3\).
Answer: B.