Simplify, $\sqrt[3]{3}\bigg(\sqrt[3]{\frac{4}{9}}-\sqrt[3]\frac{2}{9}+ \sqrt[3]\frac{1}{9}\bigg)^{-1}$.

Question

Simplify, $\sqrt[3]{3}\bigg(\sqrt[3]{\frac{4}{9}}-\sqrt[3]\frac{2}{9}+ \sqrt[3]\frac{1}{9}\bigg)^{-1}$.

What I have tried: Put $x = \frac{1}{9}$. We have, $\sqrt[3]3\big(4x^\frac{1}{3} - 2x^\frac{1}{3} + x^\frac{1}{3})$ $$\rightarrow \sqrt[3]3\big(x^\frac{1}{3}(4-2+1)\big)$$ $$\rightarrow \sqrt[3]3 \times \frac{1}{3\sqrt[3]x}$$ $$\rightarrow \frac{3}{27x} = \frac{1}{9x}$$ Putting back the value of $x$ gives the answer to be $1$.

However, in my text the answer is given to be $\sqrt[3]2 + 1$ , so where did I go wrong?

Can anyone help me?

Answer 1

: $$\sqrt[3]{4/9}-\sqrt[3]{2/9}+\sqrt[3]{1/9}=\frac{(2^{1/3}+1)(2^{2/3}-2^{1/3}+1)}{(2^{1/3}+1)(3^{2/3})}=\frac{3}{3^{2/3}(2^{1/3}+1)}$$ so:$$\sqrt[3]{3}\bigg(\sqrt[3]{\frac{4}{9}}-\sqrt[3]\frac{2}{9}+ \sqrt[3]\frac{1}{9}\bigg)^{-1}=\sqrt[3]{3}\cdot \frac{(2^{1/3}+1)(3^{2/3})}{3}=\sqrt[3]{2}+1$$