What is (2x^{-3}3y^{-2}z^{-5})^{-1} ?

The solution to (2x^{-3}3y^{-2}z^{-5})^{-1} is (x^3y^2z^5)/6

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simplify (2x^{-3}3y^{-2}z^{-5})^{-1}

simplify

(2 x^{- 3} 3 y^{- 2} z^{- 5})^{- 1}

\frac{x ^{3} y ^{2} z ^{5}}{6}

Solution steps

(2 x^{- 3} \cdot 3 y^{- 2} z^{- 5})^{- 1}

Apply exponent rule:

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

(2 x^{- 3} \cdot 3 y^{- 2} z^{- 5})^{- 1} = \frac{1}{2 x ^{- 3} \cdot 3 y ^{- 2} z ^{- 5}}

= \frac{1}{2 x ^{- 3} \cdot 3 y ^{- 2} z ^{- 5}}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{1}{6 x ^{- 3} y ^{- 2} z ^{- 5}}

Multiply

6 x^{- 3} y^{- 2} z^{- 5} : \frac{6}{x ^{3} y ^{2} z ^{5}}

= \frac{1}{\frac{6}{x ^{3} y ^{2} z ^{5}}}

Apply the fraction rule:

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

= \frac{x ^{3} y ^{2} z ^{5}}{6}

- 36 x + 13 x^{3} - x^{5}

(1 + 3 i)^{- 10}

- 6 - (- 18)

(- 4) (- 6)

\frac{45}{( - 5 )}

What is (2x^{-3}3y^{-2}z^{-5})^{-1} ?
The solution to (2x^{-3}3y^{-2}z^{-5})^{-1} is (x^3y^2z^5)/6