Answer
$$(4x^{-5}y^{2})^{-3}=\frac{x^{15}}{64y^{6}}$$
Work Step by Step
$$(4x^{-5}y^{2})^{-3}$$
Recall the power rule: $(a^{m})^{n}=a^{mn}$
Thus,
$$(4x^{-5}y^{2})^{-3}$$ $$=4^{-3}x^{-5(-3)}y^{2(-3)}$$ $$=4^{-3}x^{15}y^{-6}$$
Recall the negative exponent rule: $a^{−n}=\frac{1}{a^{n}}$ and $\frac{1}{a^{-n}} = a^{n}$
Thus,
$$=4^{-3}x^{15}y^{-6}$$ $$=\frac{x^{15}}{4^{3}y^{6}}$$ $$=\frac{x^{15}}{4^{3}y^{6}}$$ $$=\frac{x^{15}}{64y^{6}}$$
