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(three 'x' to the power of negative three, 'y' to the power of negative two) to the power of negative two

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On questions like these, I find it easiest to expand everything to be able to show my work.  We also have to remember that, since we are dealing with a negative power, that our expression will be in the form of a fraction.  So,

Step 1: 1/(3x-3·y-2)2 = 1/(3x-3·y-2)(3x-3·y-2)

once it has been expanded out

Then, we need to remember that something raised to a negative power is the same as a fraction where the negative part is on the bottom of a fraction.  This means it can be re-written as

Step 2: 1/(3/x3·1/y2)(3/x3·1/y2)

This can be simplified to:

Step 3: (1/3/x3y2)(1/3/x3y2)

Yes, this is a fraction over a fraction, but all we need to do is to treat the 1 as the numerator, and the 3/x3y2 as the denominator.  Well, we know that to divide fractions, that we must multiply by the reciprocal.  This leads us to:

Step 4: (1)(x3y2/3)·(1)(x3y2/3)

We know that to multiply exponents that we add them.  So, that leaves us with the answer:

Step 5: x6y4/9


Using properties of exponents remember that

(xn)m = x(m*n) and x-n = 1/xn

So using your equation we have:

(3x-3y-2)-2 = 3-2x(-3*-2)y(-2*-2) = (1/9)x6y4