<s>[INST] How to simply this radical expression $\dfrac{\sqrt[3]{3}}{\sqrt[3]{1}+\sqrt[3]{2}}=\sqrt[3]{\sqrt[3]{2}-1}$ $$\dfrac{\sqrt[3]{3}}{\sqrt[3]{1}+\sqrt[3]{2}}=\sqrt[3]{\sqrt[3]{2}-1}$$
I could not multiply by the conjugate since it is a cube root. Can you show me a way to simplify it?
Thanks!
 [/INST] Note $(\sqrt[3]{2}-1)(\sqrt[3]{4}+\sqrt[3]{2}+1)=(\sqrt[3]{2})^3-1=1$ and
\begin{align}
\sqrt[3]{\sqrt[3]{2}-1}
=& \sqrt[3]{\frac1{\sqrt[3]{4}+\sqrt[3]{2}+1}}
=\sqrt[3]{\frac3{2+3\sqrt[3]{4}+3\sqrt[3]{2}+1}}
=\sqrt[3]{\frac3{(\sqrt[3]{2}+1)^3}}=  
 \frac{\sqrt[3]{3}}{\sqrt[3]{2}+1}
\end{align}
</s>