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

Tutors, please sign in to answer this question.

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}