<s>[INST] How to simplify $\frac{1}{\sqrt[3]{3}-1} - \frac{2}{\sqrt[3]{3}+1}$ to $\sqrt[3]{3}$ $$\frac{1}{\sqrt[3]{3}-1} - \frac{2}{\sqrt[3]{3}+1}$$
I have simplified above to:
$$\frac{3-\sqrt[3]{3}}{(\sqrt[3]{3}+1)(\sqrt[3]{3}-1)}$$
What is equal to:
$$\frac{3-\sqrt[3]{3}}{\sqrt[3]{9}-1}$$
WolframAlpha says this can be shown as $\sqrt[3]{3}$, but I can't find out how to do this.
 [/INST] Let $t=\sqrt[3]3$. Then, we have $t^3=3$, so
$$\frac{1}{t-1}-\frac{2}{t+1}=\frac{3-t}{t^2-1}=\frac{t(3-t)}{t(t^2-1)}=\frac{t(3-t)}{3-t}=t.$$
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