Answer
$\dfrac{(a^{-2}b^{3})^{1/8}}{(a^{-3}b)^{-1/4}}=\dfrac{b^{5/8}}{a}$
Work Step by Step
$\dfrac{(a^{-2}b^{3})^{1/8}}{(a^{-3}b)^{-1/4}}$
Evaluate the powers indicated in the numerator and in the denominator:
$\dfrac{(a^{-2}b^{3})^{1/8}}{(a^{-3}b)^{-1/4}}=\dfrac{a^{-1/4}b^{3/8}}{a^{3/4}b^{-1/4}}=...$
Evaluate the division. Give the answer without negative exponents:
$...=a^{-1/4-3/4}b^{3/8+1/4}=a^{-1}b^{5/8}=\dfrac{b^{5/8}}{a}$
