How do you simplify (\frac { - 2x ^ { 3} y ^ { 0} z } { 4x z ^ { 2} } ) ^ { 5}?

1 Answer
Dec 23, 2016

$\frac{- {x}^{10}}{32 {z}^{5}}$

Explanation:

For the first simplification we can use the rule for exponents:

$\textcolor{red}{{a}^{0} = 1}$

${\left(\frac{- 2 {x}^{3} \textcolor{red}{{y}^{0}} z}{4 x {z}^{2}}\right)}^{5} \to {\left(\frac{- 2 {x}^{3} \textcolor{red}{1} z}{4 x {z}^{2}}\right)}^{5} \to {\left(\frac{- 2 {x}^{3} z}{4 x {z}^{2}}\right)}^{5}$

Next simplification is to reduce the constants:

${\left(\frac{- 2 {x}^{3} z}{4 x {z}^{2}}\right)}^{5} \to {\left(\frac{- 2 {x}^{3} z}{\left(\textcolor{red}{2 \cdot 2}\right) x {z}^{2}}\right)}^{5} \to$

${\left(\frac{- \textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} {x}^{3} z}{\left(\textcolor{red}{\cancel{2} \cdot 2}\right) x {z}^{2}}\right)}^{5} \to {\left(\frac{- {x}^{3} z}{2 x {z}^{2}}\right)}^{5}$

Next, we can take advantage of two other rules for exponents:

1. color(blue)(x^a/x^b = x^(a-b)

2. color(red)(x^a/x^b = 1/x^(b-a)

${\left(\frac{- \textcolor{b l u e}{{x}^{3}} \textcolor{red}{z}}{2 \textcolor{b l u e}{x} \textcolor{red}{{z}^{2}}}\right)}^{5} \to {\left(\frac{- \textcolor{b l u e}{{x}^{3 - 1}}}{2 \textcolor{red}{{z}^{2 - 1}}}\right)}^{5} \to {\left(\frac{- {x}^{2}}{2 {z}^{1}}\right)}^{5}$

Now, we can use yet another rule of exponents to further simplify this expression:

$\textcolor{red}{{\left({x}^{a}\right)}^{b}} = {x}^{a \cdot b}$

${\left(\frac{- {x}^{2}}{2 {z}^{1}}\right)}^{5} \to \frac{- {x}^{\textcolor{red}{2 \cdot 5}}}{{2}^{\textcolor{red}{5}} {z}^{\textcolor{red}{1 \cdot 5}}} \to \frac{- {x}^{10}}{32 {z}^{5}}$