Simplify $$\frac{1}{\sqrt[3]{2}+\sqrt[3]{16}}$$.
cube root 16 = 2cube root 2
So the expression simplifies to 1 / [3 * 3(2)]......multiply top/bottom by 3(2^2) and we have
3(2^2) / 3 * 3(2^3) =
3(2^2) / 3* 2 =
3 [ 4 ] / 6
We can write \begin{align*} \frac{1}{\sqrt[3]{2}+\sqrt[3]{16}} &= \frac{1}{\sqrt[3]{2}+\sqrt[3]{4^2}}\ &= \frac{1}{\sqrt[3]{2}+2}\ &= \frac{\sqrt[3]{2}-2}{2(\sqrt[3]{2}-2)}\ &= \boxed{\frac{1}{2}}. \end{align*}
That answer is wrong because 3sqrt4^2 is not equal to 2.