Simple math problem, just forgot to do it? Help

The answer is A, just don't know how to get it.
Thanks
Math problem in pic
http://i47.tinypic.com/1zd64on.jpg

Inside the parentheses, y^0 = 1, and x^-1 = 1/x, so the term with y goes away and the x can go in the denominator.

(3/xz^2)^(-3)

If you flip the fraction, the exponent will become positive; you'll have
(xz^2 / 3)^3

Cube everything inside the parentheses:
x^3 = x^3; (z^2)^3 = z^6; 3^3 = 27, so you're left with

[(x^3 * z^6)/27]

(3*x^-1*y^0 / z^2)^-3

First, take care of the negative exponent by flipping the fraction:
(3*x^-1*y^0 / z^2)^-3
(z^2 / 3*x^-1*y^0)^3

Get rid of y^0 since that is 1.
(z^2 / 3*x^-1)^3

Distribute the 3:
(z^2)^3 / (3^3 * (x^-1)^3)

When you raise an exponent to a power, you multiply the exponents:
(z^2)^3 / (3^3 * (x^-1)^3)
z^(2*3) / (3^3 * x^(-1*3))
z^6 / 27x^-3

Move the negative exponents to the opposite sign and multiply the exponent by -1:
z^6 * x^3 / 27

The answer is (a).

Multiply the -3 exponent to all the exponents of the values within the brackets, then simplify.

[(3 x^-1 y^0) / z^2]^-3

= [(3^-3 x^3 y^0) / z^-6]

= [(1/3^3 x^3 (1)) z^6]

= x^3 z^6 / 27

Answer is (a)