Answer
$-\frac{4}{a^{3}}$
Work Step by Step
Using the rule $\frac{x^{a}}{x^{b}}=x^{a-b}$, we obtain:
$\frac{28a^{2}b^{3}}{-7a^{5}b^{3}}$
$\frac{28}{-7}\times\frac{a^{2}}{a^{5}}\times\frac{b^{3}}{b^{3}}$
$=-4\times a^{2-5}\times b^{3-3}$
$=-4\times a^{-3}\times b^{0}$
$=-4\times \frac{1}{a^{3}}\times 1$
$=\frac{-4}{a^{3}}$
$=-\frac{4}{a^{3}}$
