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!
 A: We can use rationalization by a double step
$$\frac2{2 - \sqrt[4]{2}} \cdot \frac{2 +\sqrt[4]{2}} {2 + \sqrt[4]{2}} \cdot \frac{4+\sqrt {2}} {4+ \sqrt{2}}=\frac{(2 +\sqrt[4]{2})(4+\sqrt {2})}7$$
A: $$\frac{2}{2-\sqrt[4]2}=\frac{\sqrt[4]{8}}{\sqrt[4]{8}-1}$$ 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.
A: I would not use wolfram alpha. Rather rely on the back of the book for steps if possible. Or better yet, do it by hand first. Wolfram alpha can you give another result that's equivalent but looks different.
Typically when you see a radical in a denominator of a fraction we prefer to rationalize denominator. So in this case, multiply top and bottom by the conjugate of the denominator (same as denominator but it will have a plus instead of minus). From there distribute numerator and foil denominator (should be easy). From there simplify and if need be rationalize denominator again. Best of luck.
A: 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 $\alpha=\sqrt[4]{2}$.
