# How do you simplify ((a^-2b^4c^5)/(a^-4b^-4c^3))^2?

Jul 14, 2017

${\left(\frac{{a}^{- 2} {b}^{4} {c}^{5}}{{a}^{- 4} {b}^{- 4} {c}^{3}}\right)}^{2} = {a}^{4} {b}^{16} {c}^{4}$

#### Explanation:

Let us use the identities ${a}^{- m} = \frac{1}{a} ^ m$, $\frac{1}{a} ^ \left(- m\right) = {a}^{m}$, ${a}^{m} \times {a}^{n} = {a}^{m + n}$, ${a}^{m} / {a}^{n} = {a}^{m - n}$ and ${\left({a}^{m}\right)}^{n} = {a}^{m n}$

Hence ${\left(\frac{{a}^{- 2} {b}^{4} {c}^{5}}{{a}^{- 4} {b}^{- 4} {c}^{3}}\right)}^{2}$

= ${\left(\frac{\frac{1}{a} ^ 2 {b}^{4} {c}^{5}}{\frac{1}{a} ^ 4 \times \frac{1}{b} ^ 4 {c}^{3}}\right)}^{2}$

= ${\left(\frac{{a}^{4} {b}^{4} {b}^{4} {c}^{5}}{{a}^{2} {c}^{3}}\right)}^{2}$

= ${\left({a}^{4 - 2} {b}^{4 + 4} {c}^{5 - 3}\right)}^{2}$

= ${\left({a}^{2} {b}^{8} {c}^{2}\right)}^{2}$

= ${a}^{2 \times 2} {b}^{8 \times 2} {c}^{2 \times 2}$

= ${a}^{4} {b}^{16} {c}^{4}$