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

The solution to ((z^{-3})/(3z^{-1)})^2 is 1/(9z^4)

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

simplify

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

\frac{1}{9 z ^{4}}

Solution steps

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

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

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

Simplify

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

= \frac{\frac{1}{z ^{6}}}{( 3 z ^{- 1} ) ^{2}}

Simplify

(3 z^{- 1})^{2} : \frac{9}{z ^{2}}

= \frac{\frac{1}{z ^{6}}}{\frac{9}{z ^{2}}}

Divide fractions:

\frac{\frac{a}{b}}{\frac{c}{d}} = \frac{a \cdot d}{b \cdot c}

= \frac{1 \cdot z ^{2}}{z ^{6} \cdot 9}

Refine

= \frac{z ^{2}}{z ^{6} \cdot 9}

Apply exponent rule:

\frac{x ^{a}}{x ^{b}} = \frac{1}{x ^{b - a}}

\frac{z ^{2}}{z ^{6}} = \frac{1}{z ^{6 - 2}}

= \frac{1}{9 z ^{6 - 2}}

Subtract the numbers:

6 - 2 = 4

= \frac{1}{9 z ^{4}}

- 4 + 7

\frac{x ^{2} - 81}{x ^{4} - 81} \leq 0

\frac{1}{3} k^{3} \frac{1}{n ^{7}}

- \frac{3}{10} x - 7 < \frac{1}{2}

- 5 x + 8 = - 7

What is ((z^{-3})/(3z^{-1)})^2 ?
The solution to ((z^{-3})/(3z^{-1)})^2 is 1/(9z^4)