# How do you simplify (x^-1y^-2z^3)^-2 (x^2y^-4z^6)?

Mar 15, 2016

Use the following three properties:

${\left(x \cdot y \cdot z \cdot \ldots\right)}^{a} = {x}^{a} \cdot {y}^{a} \cdot {z}^{a} \cdot \ldots$

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

${x}^{a} \cdot {x}^{b} = {x}^{a + b}$

${x}^{4} {y}^{0} {z}^{0}$

#### Explanation:

${\left({x}^{-} 1 {y}^{-} 2 {z}^{3}\right)}^{-} 2 \left({x}^{2} {y}^{-} 4 {z}^{6}\right)$

• Use ${\left(x \cdot y \cdot z \cdot \ldots\right)}^{a} = {x}^{a} \cdot {y}^{a} \cdot {z}^{a} \cdot \ldots$

${\left({x}^{-} 1\right)}^{-} 2 {\left({y}^{-} 2\right)}^{-} 2 {\left({z}^{3}\right)}^{-} 2 {x}^{2} {y}^{-} 4 {z}^{6}$

• Use ${\left({x}^{a}\right)}^{b} = {x}^{a \cdot b}$

${x}^{- 1 \cdot \left(- 2\right)} {y}^{- 2 \cdot \left(- 2\right)} {z}^{3 \cdot \left(- 2\right)} {x}^{2} {y}^{-} 4 {z}^{6}$

${x}^{2} {y}^{4} {z}^{-} 6 {x}^{2} {y}^{-} 4 {z}^{6}$

• Use ${x}^{a} \cdot {x}^{b} = {x}^{a + b}$

${x}^{2 + 2} {y}^{4 + \left(- 4\right)} {z}^{- 6 + 6}$

${x}^{4} {y}^{0} {z}^{0}$

Note: don't say that ${y}^{0} = 1$ and ${z}^{0} = 1$ because that is not true for $y = 0$ and $z = 0$

If you wish you can use ${x}^{-} a = \frac{1}{x} ^ a$ but it's easier without it.