#### Answer

$\frac{1}{125a^9b^3c^{21}}$

#### Work Step by Step

Use the outside exponent on each number within the parentheses. Remember to multiply the exponents of the bases by the outside exponent. In this case, we are multiplying $3$ by $-3$, $1$ by $-3,$ and $7$ by $-3$ for $a$, $b$, and $c$, respectively. Move any resulting negative exponents to the denominator in order to turn them positive, and leave any resulting numbers with positive exponents to the numerator (in this case, $1$).
$(5a^3bc^7)^{-3}$
$5^{-3}a^{-9}b^{-3}c^{-21}$
$\frac{1}{125a^9b^3c^{21}}$