Answer
$$-\frac{81a^{18}}{16b^{32}c^{4}}$$
Work Step by Step
$$(\frac{-2a^{-4}b^{3}c^{-1}}{3a^{-2}b^{-5}c^{-2}})^{-4}$$
Recall the fraction rule: $\frac{-a}{b} = -\frac{a}{b}$
Thus,
$=(\frac{-2a^{-4}b^{3}c^{-1}}{3a^{-2}b^{-5}c^{-2}})^{-4} = (-\frac{2a^{-4}b^{3}c^{-1}}{3a^{-2}b^{-5}c^{-2}})^{-4}$
Recall the power rule: $(a^{m})^{n}=a^{mn}$
Thus,
$=(-\frac{2a^{-4}b^{3}c^{-1}}{3a^{-2}b^{-5}c^{-2}})^{-4}$
$=-\frac{2^{-4}a^{-4\cdot-4}b^{3\cdot-4}c^{-1\cdot-4}}{3^{-4}a^{-2\cdot-4}b^{-5\cdot-4}c^{-2\cdot-4}}$
$=-\frac{2^{-4}a^{16}b^{-12}c^{4}}{3^{-4}a^{8}b^{20}c^{8}}$
Recall the negative exponent rule: $a^{−n}=\frac{1}{a^{n}}$ and $\frac{1}{a^{-n}} = a^{n}$
Thus,
$=-\frac{2^{-4}a^{16}b^{-12}c^{4}}{3^{-4}a^{8}b^{20}c^{8}}$
$=-\frac{3^{4}a^{16}b^{-12}c^{4}}{2^{4}a^{8}b^{20}c^{8}}$
Recall the quotient rule: $\frac{a^{m}}{a^{n}}=a^{m-n}$ and $\frac{a^{n}}{a^{m}}=\frac{1}{a^{m+n}}$ if $m>n$
$=-\frac{3^{4}a^{16}b^{-12}c^{4}}{2^{4}a^{8}b^{20}c^{8}}$
$=-\frac{3^{4}a^{16-8}}{2^{4}b^{20+12}c^{8-4}}$
$=-\frac{81a^{18}}{16b^{32}c^{4}}$
