# How do you simplify (8^-2z^-3y)^-1/(5y^2z^-2)^3(5yz^-2)^-1?

Apr 14, 2016

$\frac{64}{625} \times {z}^{11} / {y}^{8}$

#### Explanation:

Take it one step at a time! break the question down into parts. Solve each part then put it all back together again.
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$\textcolor{b l u e}{\text{Consider } \left({8}^{- 2} {z}^{- 3} y\right)}$

$\implies \frac{1}{8} ^ 2 \times \frac{1}{z} ^ 3 \times \frac{y}{1} = \frac{y}{{8}^{2} {z}^{3}}$

But the whole thing is: ${\left({8}^{- 2} {z}^{- 3} y\right)}^{- 1}$

So ${\left(\frac{y}{{8}^{2} {z}^{3}}\right)}^{- 1} \textcolor{g r e e n}{\to \frac{{8}^{2} {z}^{3}}{y}}$
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$\textcolor{b l u e}{\text{Consider } {\left(5 {y}^{2} {z}^{- 2}\right)}^{3}}$

$\implies \frac{5 {y}^{2}}{1} \times \frac{1}{z} ^ 2 = \frac{5 {y}^{2}}{z} ^ 2$

But the whole thing is ${\left(\frac{5 {y}^{2}}{z} ^ 2\right)}^{3}$

$\implies \textcolor{g r e e n}{\frac{{5}^{3} {y}^{6}}{z} ^ 6} \text{ }$ ( divide the whole by this one!)
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color(blue)("Consider "(5yz^(-2))^(-1)

$5 y {z}^{- 2} \to \frac{5 y}{z} ^ 2$

But the whole thing is ${\left(\frac{5 y}{z} ^ 2\right)}^{- 1}$

$\textcolor{g r e e n}{\implies \frac{{z}^{2}}{5 y}}$
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$\textcolor{b l u e}{\text{Putting at all together}}$

$\frac{{8}^{2} {z}^{3}}{y} \times \frac{{z}^{2}}{5 y} \div \frac{{5}^{3} {y}^{6}}{z} ^ 6$

$\frac{{8}^{2} {z}^{3}}{y} \times \frac{{z}^{2}}{5 y} \times \frac{{z}^{6}}{{5}^{3} {y}^{6}}$

$\frac{64}{625} \times {z}^{11} / {y}^{8}$