What is (2x^4y^{(-4)/5})(8y^2)^{2/3} ?

The solution to (2x^4y^{(-4)/5})(8y^2)^{2/3} is 8x^4y^{8/15}

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simplify (2x^4y^{(-4)/5})(8y^2)^{2/3}

simplify

(2 x^{4} y^{\frac{- 4}{5}}) (8 y^{2})^{\frac{2}{3}}

8 x^{4} y^{\frac{8}{15}}

Solution steps

(2 x^{4} y^{\frac{- 4}{5}}) (8 y^{2})^{\frac{2}{3}}

Simplify

2 x^{4} y^{\frac{- 4}{5}} : \frac{2 x ^{4}}{y ^{\frac{4}{5}}}

= \frac{2 x ^{4}}{y ^{\frac{4}{5}}} (8 y^{2})^{\frac{2}{3}}

Simplify

(8 y^{2})^{\frac{2}{3}} : 4 y^{\frac{4}{3}}

= \frac{2 x ^{4}}{y ^{\frac{4}{5}}} \cdot 4 y^{\frac{4}{3}}

Apply the fraction rule:

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

= \frac{2 x ^{4} \cdot 4 y ^{\frac{4}{3}}}{y ^{\frac{4}{5}}}

Simplify

\frac{y ^{\frac{4}{3}}}{y ^{\frac{4}{5}}} : y^{\frac{8}{15}}

= 2 x^{4} \cdot 4 y^{\frac{8}{15}}

Multiply the numbers:

2 \cdot 4 = 8

= 8 x^{4} y^{\frac{8}{15}}

\frac{4 + i}{4 - i}

- 3 ∣ 1 - 4 q ∣ - 4 \leq - 14

- X^{2} \cdot X

64 x^{2} + 80 x + 25 = 0

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

What is (2x^4y^{(-4)/5})(8y^2)^{2/3} ?
The solution to (2x^4y^{(-4)/5})(8y^2)^{2/3} is 8x^4y^{8/15}