<s>[INST] Rationalize the denominator of $2\over{2 - \sqrt[4]{2}}$? 
Rationalize the denominator of $2\over{2 - \sqrt[4]{2}}$.

Here's my progress. Let $x = \sqrt[4]{2}$. Then our expression can be written as $x^4/(x^4 - x)$, which simplifies to $x^3/(x^3 - 1)$. Multiply top and bottom by $(x^3 + 1)$ to get $x^3(x^3 + 1)/(x^6 - 1)$. Multiply top and bottom by $(x^6 + 1)$ to get$$x^3(x^3 + 1)(x^6 + 1)/(x^{12} - 1) = x^3(x^3 + 1)(x^6 + 1)/7 = {1\over7}(8 + 4\sqrt[4]{2} + 2 \sqrt{2} + 2^{3/4}).$$However, Wolfram Alpha also tells me that we can write this as$${1\over{14}}\Big(16 + 4\sqrt{2} + 7\sqrt{{{64}\over{49}} + {{72{\sqrt2}}\over{49}}}\Big)$$But how do I derive that? Seems impossible!
 [/INST] $$\frac{2}{2-\sqrt[4]2}=\frac{\sqrt[4]{8}}{\sqrt[4]{8}-1}$$ and use $$(x-1)(x+1)(x^2+1)=x^4-1$$ for $x=\sqrt[4]{8}.$
Now, about the WA's bang.
$$7\sqrt{\frac{64}{49}+\frac{72\sqrt2}{49}}=\sqrt{64+72\sqrt2}=\sqrt{8\sqrt2(4\sqrt2+9)}=$$
$$=2\sqrt[4]8\sqrt{(2\sqrt2)^2+4\sqrt2+1}=2\sqrt[4]8(2\sqrt2+1)$$ and we made denesting.
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