<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 $\sqrt[6]2=x\implies\sqrt[3]2=x^2$
We have $$\dfrac1{2+2x+2x^2}=\dfrac{1-x}{2(1-x^3)}=\dfrac{(1-x)(1+x^3)}{2(1-x^6)}$$
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