<s>[INST] Rationalize $\frac{1}{2 + 2\sqrt[6]{2} + 2\sqrt[3]{2}}$ I am having trouble rationalizing the denominator of $$\frac{1}{2 + 2\sqrt[6]{2} + 2\sqrt[3]{2}}$$
I tried grouping the denominator as $(2 + 2\sqrt[6]{2}) + 2\sqrt[3]{2}$ and multiplying top and bottom by $(2 + 2\sqrt[6]{2})^2-(2 + 2\sqrt[6]{2})(2\sqrt[3]{2})+(2\sqrt[3]{2})^2$ to obtain  $$\frac{(2+2\sqrt[6]{2})^2-(2 + 2\sqrt[6]{2})(2\sqrt[3]{2})+(2\sqrt[3]{2})^2}{(2+2\sqrt[6]{2})^3+16}$$
However, expanding the denominator produces $24 + 24\sqrt[6]{2}+24\sqrt[3]{2}+8\sqrt{2}$ which doesn't look any better than the original.
So what should I do differently or is there another way to approach this problem? Any advice would be appreciated. Thanks.
 [/INST] Let $a=\sqrt[6]2$. Then your expression is$$\frac1{2+2a+2a^2}=\frac{1-a}{2(1+a+a^2)(1-a)}=\frac{1-a}{2(1-a^3)}=\frac{1-\sqrt[6]2}{2\bigl(1-\sqrt2\bigr)}=\frac{\bigl(1-\sqrt[6]2\bigr)\bigl(1+\sqrt2\bigr)}{-2}.$$
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