What is (\sqrt[3]{1+2q+q^2})^2 ?

The solution to (\sqrt[3]{1+2q+q^2})^2 is (q^2+2q+1)^{2/3}

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simplify (\sqrt[3]{1+2q+q^2})^2

simplify

(q^{2} + 2 q + 1)^{\frac{2}{3}}

Solution steps

Apply radical rule:

= ((1 + 2 q + q^{2})^{\frac{1}{3}})^{2}

Apply exponent rule:

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

= (1 + 2 q + q^{2})^{\frac{1}{3} \cdot 2}

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

= (1 + 2 q + q^{2})^{\frac{2}{3}}

= (q^{2} + 2 q + 1)^{\frac{2}{3}}

2 e^{- l n (8)}

- \frac{1}{2} e^{- x} (e^{2 x} - 1)

(2 16 - x^{2}) (x 16 - x^{2})

\frac{23}{2}

What is (\sqrt[3]{1+2q+q^2})^2 ?
The solution to (\sqrt[3]{1+2q+q^2})^2 is (q^2+2q+1)^{2/3}