Answer
$$ \frac{y^{5}}{8x^{14}}$$
Work Step by Step
$$(-4x^{3}y^{-5})^{-2}(2x^{-8}y^{-5})$$
Simplify the first term $(-4x^{3}y^{-5})^{-2}$
Recall the power rule: $(a^{m})^{n}=a^{mn}$
Thus,
$$(-4x^{3}y^{-5})^{-2} = -4^{-2}x^{3(-2)}y^{-5(-2)}=-4^{-2}x^{-6}y^{10}$$
Rewrite the equation:
$$(-4^{-2}x^{-6}y^{10})(2x^{-8}y^{-5})$$
Recall the product rule: $a^{m}⋅a^{n}=a^{m+n}$
Thus,
$$(-4^{-2}x^{-6}y^{10})(2x^{-8}y^{-5}) = (-4^{-2})(2)(x^{-6-8}y^{10-5}) = (-4^{-2})(2)(x^{-14}y^{5})$$
Recall the negative exponent rule: $a^{−n}=\frac{1}{a^{n}}$ and $\frac{1}{a^{-n}} = a^{n}$
Thus,
$$(-4^{-2})(2)(x^{-14}y^{5}) = \frac{(2)(y^{5})}{-4^{2}x^{14}}$$
$$= \frac{(2)(y^{5})}{16x^{14}}$$
$$= \frac{y^{5}}{8x^{14}}$$
