# How do you simplify (14f^-3g^2h^-7)/(21k^3)?

Mar 2, 2017

$\frac{2 {f}^{-} 3 {g}^{2} {h}^{-} 7}{3 {k}^{3}} = \frac{2 {g}^{2}}{3 {f}^{3} {h}^{7} {k}^{3}}$

#### Explanation:

Technically the only thing needed to simplify this expression is to change $\frac{14}{21}$ to $\frac{2}{3}$.

So $\frac{2 {f}^{-} 3 {g}^{2} {h}^{-} 7}{3 {k}^{3}}$ is simplified.

However, usually mathematicians prefer to have all of the exponents positive. To make them positive, you flip them to the other side of the dividing line using the exponential rules:
$\frac{1}{x} ^ - m = {x}^{m}$ and ${x}^{-} n = \frac{1}{x} ^ n$

Which leaves: $\frac{2 {g}^{2}}{3 {f}^{3} {h}^{7} {k}^{3}}$