# Did I simplify this right?

## ${\left(\frac{- 2 {a}^{3} {b}^{2}}{3 {a}^{2} {b}^{-} 3}\right)}^{-} 2$ = $\frac{- 1}{\frac{2 a}{3 b}}$

Apr 19, 2018

No, but don't let it get to you! The only way to learn is by making mistakes and lots of them!

#### Explanation:

First what you should do is get rid of the negative in the exponent. ${\left(\frac{- 2 {a}^{3} {b}^{2}}{3 {a}^{2} {b}^{- 3}}\right)}^{- 2} = {\left(\frac{3 {a}^{2} {b}^{- 3}}{- 2 {a}^{3} {b}^{2}}\right)}^{2}$

I would then advise cancelling out some of the variables in the equation. The a variables cancel out to ${a}^{-} 1$ while the b variables cancel out to ${b}^{-} 5$. Rewritten, the equation is now

${\left(\frac{3}{- 2 a {b}^{5}}\right)}^{2}$

Next you should distribute the exponent separately to both the numerator and the denominator.

${\left(3\right)}^{2} / {\left(- 2 a {b}^{5}\right)}^{2}$

Simplifying this gives

$\frac{9}{4 {a}^{2} {b}^{10}}$

What I think you might have done is simplify the fraction first (Which is completely fine) to get

${\left(\frac{- 2 a}{3 b}\right)}^{-} 2$

The thing that got you was probably the negative exponent on the b. When a variable has a negative exponent, it really means that it belongs on the top of the fraction. ${a}^{2} / {b}^{-} 2$ should really be written as ${a}^{2} \cdot {b}^{2}$

The correct simplification for this step would be

${\left(\frac{- 2 a {b}^{5}}{3}\right)}^{-} 2$

Also note that since everything is squared, you can remove the negative in front of the 2. It looks like you might have kept it outside.

Great job and keep up the good work!