What is (3x^{-3}y^3)/(9x^2(y^{-3))^2} ?

The solution to (3x^{-3}y^3)/(9x^2(y^{-3))^2} is (y^9)/(3x^5)

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simplify (3x^{-3}y^3)/(9x^2(y^{-3))^2}

simplify

\frac{3 x ^{- 3} y ^{3}}{9 x ^{2} ( y ^{- 3} ) ^{2}}

\frac{y ^{9}}{3 x ^{5}}

Solution steps

\frac{3 x ^{- 3} y ^{3}}{9 x ^{2} ( y ^{- 3} ) ^{2}}

Simplify

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

= \frac{3 x ^{- 3} y ^{3}}{9 x ^{2} \frac{1}{y ^{6}}}

Factor the number:

9 = 3 \cdot 3

= \frac{3 x ^{- 3} y ^{3}}{3 \cdot 3 x ^{2} \frac{1}{y ^{6}}}

Cancel the common factor:

3

= \frac{x ^{- 3} y ^{3}}{3 x ^{2} \frac{1}{y ^{6}}}

Simplify

\frac{x ^{- 3}}{x ^{2}} : \frac{1}{x ^{5}}

= \frac{y ^{3}}{3 x ^{5} \frac{1}{y ^{6}}}

Simplify

\frac{y ^{3}}{3 x ^{5} \frac{1}{y ^{6}}} : \frac{y ^{9}}{3 x ^{5}}

= \frac{y ^{9}}{3 x ^{5}}

x^{2} - 1 \leq 3

\frac{a ^{- 3} b ^{4}}{a ^{- 5} b ^{5}}

4 x - 3 = 3 x - 1

\frac{4}{3} - \frac{6}{5} + \frac{6}{15}

x^{2} + 3 x + 2 \geq 0

What is (3x^{-3}y^3)/(9x^2(y^{-3))^2} ?
The solution to (3x^{-3}y^3)/(9x^2(y^{-3))^2} is (y^9)/(3x^5)