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

The solution to (2x^4y^{-4/5})^3(8y^2)^{2/3} is (32x^{12})/(y^{16/15)}

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

simplify

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

\frac{32 x ^{12}}{y ^{\frac{16}{15}}}

Solution steps

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

Simplify

(2 x^{4} y^{- \frac{4}{5}})^{3} : \frac{8 x ^{12}}{y ^{\frac{12}{5}}}

= \frac{8 x ^{12}}{y ^{\frac{12}{5}}} (8 y^{2})^{\frac{2}{3}}

Simplify

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

= \frac{8 x ^{12}}{y ^{\frac{12}{5}}} \cdot 4 y^{\frac{4}{3}}

Apply the fraction rule:

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

= \frac{8 x ^{12} \cdot 4 y ^{\frac{4}{3}}}{y ^{\frac{12}{5}}}

Simplify

\frac{y ^{\frac{4}{3}}}{y ^{\frac{12}{5}}} : \frac{1}{y ^{\frac{16}{15}}}

= \frac{8 x ^{12} \cdot 4}{y ^{\frac{16}{15}}}

Multiply the numbers:

8 \cdot 4 = 32

= \frac{32 x ^{12}}{y ^{\frac{16}{15}}}

3 (2 x + 1) > 5 - 2 x

- \frac{5}{4} (- 4)

9 (2 - 3 x) - 29 = 8 x - (x - 23)

\frac{\frac{a}{25} - \frac{a}{5}}{a}

x^{2} - 9 x + 8

What is (2x^4y^{-4/5})^3(8y^2)^{2/3} ?
The solution to (2x^4y^{-4/5})^3(8y^2)^{2/3} is (32x^{12})/(y^{16/15)}