## Intermediate Algebra (6th Edition)

$\frac{1}{a^{\frac{9}{2}}}$
We are given the expression $(a^{\frac{1}{2}}a^{-2})^{3}$. First, we can use the product rule to simplify, which holds that $a^{m}\times a^{n}=a^{m+n}$ (where a is a real number, and m and n are positive integers). $(a^{\frac{1}{2}+(-2)})^{3}=(a^{-\frac{3}{2}})^{3}$ Next, we can use the power rule, which holds that $(a^{m})^{n}=a^{m\times n}$ (where a is a real number, and m and n are integers). $(a^{-\frac{3}{2}})^{3}=a^{-\frac{3}{2}\times3}=a^{-\frac{9}{2}}=\frac{1}{a^{\frac{9}{2}}}$