Question

Help me please! If you DO decide to help, please explain it to me since I'm extremely confused! THANK YOU!

Answer 1

$\bf \cfrac{(-4x^2)(2x^{-2}y)^3}{(16x^5)(4y^3)^2}\implies \cfrac{(-4x^2)(2^3x^{-2\cdot 3}y^3)}{(16x^5)(4^2y^{3\cdot 2})}\implies \cfrac{(-4x^2)(8x^{-6}y^3)}{(16x^5)(16y^6)} \\\\\\ \cfrac{-32x^{2-6}y^3}{256x^5y^6}\implies -\cfrac{x^{-4} y^3}{8x^5y^6}\implies -\cfrac{1}{8x^5x^{4}y^6y^{-3}}\implies -\cfrac{1}{8x^{5+4}y^{6-3}} \\\\\\ -\cfrac{1}{8x^9y^3}$

Answer 2

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: $(2x^{-2}y)^3=(2^{1*3}x^{-2*3}y^{1*3})=(8x^{-6}y^3)$ and $(4y^3)^2=(4^{1*2}y^{3*2})=(16y^6)$

Our current equation look like this: $\frac{(-4x^2)(8x^{-6}y^3)}{(16x^5)(16y^6)}$

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: $\frac{(-4x^2)(8x^{-6}y^3)}{(16x^5)(16y^6)}=\frac{-32x^{2+(-6)}y^3}{256x^5y^6} =\frac{-32x^{-4}y^3}{256x^5y^6}$

Now we can divide. The rule with dividing exponents with the same base is that you subtract them.

In this case: $\frac{-32x^{-4}y^3}{256x^5y^6}=-\frac{1}{8} x^{-4-5}y^{3-6}=-\frac{1}{8} x^{-9}y^{-3}$

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: $-\frac{1}{8x^9y^3}$ . And that would be your final answer.