<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] $$\frac{3-\sqrt[3]{3}}{\sqrt[3]{9}-1}=\frac{(3^{1/3})^3-3^{1/3}}{3^{2/3}-1}=\frac{3^{1/3}(3^{2/3}-1)}{3^{2/3}-1}=3^{1/3}(=\sqrt[3]{3})$$
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