What is 1/((2y^2z)^{-3)} ?

The solution to 1/((2y^2z)^{-3)} is 8y^6z^3

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simplify 1/((2y^2z)^{-3)}

simplify

\frac{1}{( 2 y ^{2} z ) ^{- 3}}

8 y^{6} z^{3}

Solution steps

\frac{1}{( 2 y ^{2} z ) ^{- 3}}

Simplify

(2 y^{2} z)^{- 3} : \frac{1}{8 y ^{6} z ^{3}}

= \frac{1}{\frac{1}{8 y ^{6} z ^{3}}}

Apply the fraction rule:

\frac{1}{\frac{b}{c}} = \frac{c}{b}

= \frac{8 y ^{6} z ^{3}}{1}

Apply the fraction rule:

\frac{a}{1} = a

= 8 y^{6} z^{3}

\frac{1}{( 1 + i ) ^{2}}

- 9 \cdot (- 6)

(6 v^{2} + 2 v - 9) (4 - 5 v)

x - \frac{1}{x}