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!
 A: $a^3 + b^3 = (a+b)(a^2 - ab + b^2)$
Since you want to clear the denominator, try multiplying both numerator and denominator by $(2^{\frac{2}{3}} - 2^{\frac{1}{3}} + 1)$
That will give you 2-1 for the denominator.
A: 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. $\blacksquare$
A: 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}
