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!

Cubing both sides, we wish to prove:
$$\sqrt[3]{2}-1 = \bigg(\frac{\sqrt[3]{3}}{1 + \sqrt[3]{2}}\bigg)^{3}$$
We have:
$$\bigg(\frac{\sqrt[3]{3}}{1 + \sqrt[3]{2}}\bigg)^{3} = \frac{3}{1 + 3\sqrt[3]{2} + 3\sqrt[3]{2}^{2} + 2}=\frac{1}{1+\sqrt[3]{2}+\sqrt[3]{4}}=\frac{1}{1+1\cdot\sqrt[3]{2} + \sqrt[3]{2}^{2}}\cdot\frac{\sqrt[3]{2}-1}{\sqrt[3]{2}-1}$$
Because $(a - b)(a^{2} + ab + b^{2}) = a^{3} - b^{3}$, this simplifies to:
$$=\frac{\sqrt[3]{2}-1}{\sqrt[3]{2}^{3} - 1} = \frac{\sqrt[3]{2}-1}{2-1}=\sqrt[3]{2}-1$$
Thus, this identity is true. $\blacksquare$
