What is (\sqrt[3]{x^2+3x})^{4=16} ?

The solution to (\sqrt[3]{x^2+3x})^{4=16} is (x^2+3x)^{16*4/3 =}

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simplify (\sqrt[3]{x^2+3x})^{4=16}

simplify

(x^{2} + 3 x)^{16 \cdot \frac{4}{3} =}

Solution steps

Apply radical rule:

= ((x^{2} + 3 x)^{\frac{1}{3}})^{4 = 16}

Apply exponent rule:

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

assuming

a \geq 0

= (x^{2} + 3 x)^{\frac{1}{3} \cdot 4 = 16}

\frac{1}{3} \cdot 4 = \frac{4}{3}

= (x^{2} + 3 x)^{16 \cdot \frac{4}{3} =}

\frac{2. 5 ^{- 4}}{7. 5 ^{- 4}}

(4 b^{3}) (8 b^{2})

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What is (\sqrt[3]{x^2+3x})^{4=16} ?
The solution to (\sqrt[3]{x^2+3x})^{4=16} is (x^2+3x)^{16*4/3 =}