What is 3y^{-4/3}*2\sqrt[3]{y} ?

The solution to 3y^{-4/3}*2\sqrt[3]{y} is 6/y

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simplify 3y^{-4/3}*2\sqrt[3]{y}

simplify

\frac{6}{y}

Solution steps

Apply radical rule:

= 3 y^{- \frac{4}{3}} \cdot 2 y^{\frac{1}{3}}

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

y^{- \frac{4}{3}} y^{\frac{1}{3}} = y^{- \frac{4}{3} + \frac{1}{3}}

= 3 y^{- \frac{4}{3} + \frac{1}{3}} \cdot 2

- \frac{4}{3} + \frac{1}{3} = - 1

= 3 y^{- 1} \cdot 2

Apply exponent rule:

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

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

= 3 \cdot \frac{1}{y} \cdot 2

Multiply the numbers:

3 \cdot 2 = 6

= 6 \cdot \frac{1}{y}

Apply the fraction rule:

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

= \frac{6 \cdot 1}{y}

Multiply the numbers:

6 \cdot 1 = 6

= \frac{6}{y}

64 y^{4} - x^{3} y^{4}

\frac{4}{x - 3} < x + 1

11 \cdot \frac{1}{2}

- 2 (x - 1)^{2} = 36

\frac{2 ( 9 ) ^{\frac{3}{2}}}{3}