Answer
$\dfrac{a^{-3}b^8}{c^{-2}} =\dfrac{b^8c^2}{a^3}$
Work Step by Step
RECALL:
The negative-exponent rule of exponents states that:
$a^{-m} =\dfrac{1}{a^m}$ and $\dfrac{1}{a^{-m}} = a^m$, where $a \ne 0$.
Use the negative-exponent rule to find:
$\dfrac{a^{-3}b^8}{c^{-2}} =\dfrac{\frac{b^8}{a^3}}{\frac{1}{c^2}}=\dfrac{b^8}{a^3} \cdot \dfrac{c^2}{1}=\dfrac{b^8c^2}{a^3}$
