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!

The systematic way is to compute the inverse of $2-x$ mod $x^4-2$ using the extended Euclidean algorithm:
$$
1= \frac{1}{14}(x^4-2) + \frac{1}{14} (x^3 + 2 x^2 + 4 x + 8)(2-x)
$$
(courtesy of WA).
Therefore,
$$
\frac{1}{2 - \alpha} = \frac{1}{14} (\alpha^3 + 2 \alpha^2 + 4 \alpha + 8)
$$
where $\alpha=\sqrt[4]{2}$.
