So firstly, you want to get rid of the outermost powers. The rule with powering something with an exponent is to multiply the exponents together.
In this case: [tex] (2x^{-2}y)^3=(2^{1*3}x^{-2*3}y^{1*3})=(8x^{-6}y^3) [/tex] and [tex] (4y^3)^2=(4^{1*2}y^{3*2})=(16y^6) [/tex]
Our current equation look like this: [tex] \frac{(-4x^2)(8x^{-6}y^3)}{(16x^5)(16y^6)} [/tex]
Next, you want to multiply what's on the numerators and the denominator. The rule with multiplying exponents of the same base is to add them together.
In this case: [tex] \frac{(-4x^2)(8x^{-6}y^3)}{(16x^5)(16y^6)}=\frac{-32x^{2+(-6)}y^3}{256x^5y^6} =\frac{-32x^{-4}y^3}{256x^5y^6} [/tex]
Now we can divide. The rule with dividing exponents with the same base is that you subtract them.
In this case: [tex] \frac{-32x^{-4}y^3}{256x^5y^6}=-\frac{1}{8} x^{-4-5}y^{3-6}=-\frac{1}{8} x^{-9}y^{-3} [/tex]
Now, if you're required to leave no negative exponents, the rule is that for example x^-2 would turn into 1/x^2. In this case, it would turn like this: [tex] -\frac{1}{8x^9y^3} [/tex] . And that would be your final answer.
