<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] We can use rationalization by a double step
$$\frac2{2 - \sqrt[4]{2}} \cdot \frac{2 +\sqrt[4]{2}} {2 + \sqrt[4]{2}} \cdot \frac{4+\sqrt {2}} {4+ \sqrt{2}}=\frac{(2 +\sqrt[4]{2})(4+\sqrt {2})}7$$
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