# How do you simplify (\frac { 2x ^ { 4} y ^ { 0} z ^ { - 1} } { x ^ { - 1} y ^ { 2} \cdot x y ^ { - 4} z ^ { 4} } ) ^ { - 4}?

Feb 2, 2017

${z}^{20} / \left(16 {x}^{16} {y}^{8}\right)$

#### Explanation:

$\textcolor{b r o w n}{\text{The trick with these is to take it very slowly and 1 step at a time}}$

Note that ${y}^{0} = 1$

$\textcolor{b l u e}{\text{Consider the numerator}}$

$2 {x}^{4} \times 1 \times \frac{1}{z} = \frac{2 {x}^{4}}{z}$

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$\textcolor{b l u e}{\text{Consider the denominator}}$

$\frac{1}{x} \times {y}^{2} \times x \times \frac{1}{y} ^ 4 \times {z}^{4}$

But in fact this is:

$\frac{1}{\frac{1}{x} \times {y}^{2} \times x \times \frac{1}{y} ^ 4 \times {z}^{4}} \text{ } = \cancel{x} \times \frac{1}{\cancel{x}} \times \frac{1}{\cancel{{y}^{2}}} \times {y}^{\cancel{\textcolor{w h i t e}{.} 4} 2} \times \frac{1}{z} ^ 4$

$= {y}^{2} / {z}^{4}$

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$\textcolor{b l u e}{\text{Putting it all back together}}$

Putting the numerator and denominator back together we have:

$\frac{2 {x}^{4}}{z} \times {y}^{2} / {z}^{4} = \frac{2 {x}^{4} {y}^{2}}{z} ^ 5$

But the whole thing is raised to the power of negative 4 giving

${\left(\frac{2 {x}^{4} {y}^{2}}{{z}^{5}}\right)}^{- 4}$

${\left(\frac{{z}^{5}}{2 {x}^{4} {y}^{2}}\right)}^{4}$

${z}^{20} / \left(16 {x}^{16} {y}^{8}\right)$