What is \sqrt[3]{(x^4y^{-1})}^6 ?

The solution to \sqrt[3]{(x^4y^{-1})}^6 is (x^8)/(y^2)

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simplify \sqrt[3]{(x^4y^{-1})}^6

simplify

\frac{x ^{8}}{y ^{2}}

Solution steps

Apply rule:

(a) = a

(x^{4} y^{- 1}) = x^{4} y^{- 1}

Rewrite

6

3 \cdot 2

Apply exponent rule:

a^{b c} = (a^{b})^{c},

assuming

a \geq 0

Apply exponent rule:

assuming

n i s o dd or a \geq 0

= (x^{4} y^{- 1})^{2}

Apply exponent rule:

(a \cdot b)^{n} = a^{n} b^{n}

(x^{4} y^{- 1})^{2} = (x^{4})^{2} (y^{- 1})^{2}

= (x^{4})^{2} (y^{- 1})^{2}

Simplify

(x^{4})^{2} : x^{8}

= x^{8} (y^{- 1})^{2}

Simplify

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

= x^{8} \frac{1}{y ^{2}}

Simplify

x^{8} \frac{1}{y ^{2}} : \frac{x ^{8}}{y ^{2}}

= \frac{x ^{8}}{y ^{2}}

(7 b)^{3}

4 \cdot π \cdot 1 \cdot 1 0^{3} \cdot 22 \cdot 1 0^{- 9} \cdot 47 \cdot 1 0^{- 9}

9.32 \times 1 0^{- 10}

e^{\frac{1}{2}} \cdot 2 e^{2} \cdot 5 e^{3}

What is \sqrt[3]{(x^4y^{-1})}^6 ?
The solution to \sqrt[3]{(x^4y^{-1})}^6 is (x^8)/(y^2)