Please solve!

(3x^{-3}·y^{-2})^{-2}

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

Please solve!

(3x^{-3}·y^{-2})^{-2}

(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/x^{3}·1/y^{2})(3/x^{3}·1/y^{2})

This can be simplified to:

Step 3: (1/3/x^{3}y^{2})(1/3/x^{3}y^{2})

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/x^{3}y^{2} as the denominator. Well, we know that to divide fractions, that we must multiply by the reciprocal. This leads us to:

Step 4: (1)(x^{3}y^{2}/3)·(1)(x^{3}y^{2}/3)

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

Step 5: x^{6}y^{4}/9

Using properties of exponents remember that

(x^{n})^{m} = x^{(m*n)} and x^{-n} = 1/x^{n}

So using your equation we have:

(3x^{-3}y^{-2})^{-2} = 3^{-2}x^{(-3*-2)}y^{(-2*-2)} = (1/9)x^{6}y^{4}

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