# How do you simplify ((-15(ab)^2)/(5a^2b))^5?

Mar 7, 2016

${\left(- 3 b\right)}^{5}$ or $- 243 {b}^{5}$

#### Explanation:

So we have

${\left(\frac{- 15 {\left(a b\right)}^{2}}{5 {a}^{2} b}\right)}^{5}$

First let's take that first square in the first term, since it's a multiplication the exponent goes to both numbers.

${\left(\frac{- 15 {a}^{2} {b}^{2}}{5 {a}^{2} b}\right)}^{5}$

Now we can get rid of any letters that show up equally in top and bottom, we have 2 $a$s on top and 2 in the bottom so they both go, but we only 1 $b$ in the bottom and two on the top, so one of the $b$s on the top will remain.

${\left(\frac{- 15 b}{5}\right)}^{5}$

Now, we simplify the actual numbers, we know $15 = 3 \cdot 5$ so it's just a matter of factorizing and discarding equal factors on top and bottom.

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

Which is easy to compute if you so wish, the letter just gets the exponent, the minus sign continues to be a minus because $\left(- 1\right) \setminus \cdot {\left(- 1\right)}^{2} \cdot {\left(- 1\right)}^{2} = \left(- 1\right) \cdot 1 \cdot 1 = - 1$, which only leaves us to deal with the $3$, but since ${3}^{5} = 3 \cdot {\left(3\right)}^{2} \cdot {\left(3\right)}^{2} = 3 \cdot 9 \cdot 9 = 3 \cdot 81$, we only need to do one multiplication that isn't tabled.

$3 \cdot 81 = 3 \cdot 80 + 3 \cdot 1 = 243$, so the final expression is

${\left(- 3 b\right)}^{5}$ or $- 243 {b}^{5}$