<s>[INST] Prove that $\frac{1}{\sqrt[3]2}=\sqrt{\frac 5{\sqrt[3]4}-1}-\sqrt{(3-\sqrt[3]2)(\sqrt[3]2-1)}$ Playing around with denesting radicals, I arrived at the following formula which appears to be correct.

$$\frac 1{\sqrt[3]2}=\sqrt{\frac 5{\sqrt[3]4}-1}-\sqrt{(3-\sqrt[3]2)(\sqrt[3]2-1)}$$

If one were to prove this strictly from the given equation, say, as a contest math problem, how would one do it? I have literally no idea how to do this, and I only derive these nested radical equations backwards (e.g. substituting radical values for $a$, $b$ and $c$ in an expression like $(a+b-c)^2$ and hoping for an elegant result after some more or less tedious algebra).
Is there an official method by which to prove this, or is it a bit foggy? I have heard Galois theory is probably important here but that's all I know about it, pretty much, and the rest is vaguely known to me. I would love to see if there is some kind of process to solve/prove such problems, as it might shed light on how Ramanujan came across his several radical denestations and related general identities.

How it was discovered.
I noticed that $$1-\frac 1{\sqrt[3]2}+\frac 1{\sqrt[3]4}=\frac 12\Big\{1+\sqrt{(3-\sqrt[3]2)(\sqrt[3]2-1)}\Big\}$$ and $$1-\frac 1{2\sqrt[3]2}+\frac 1{\sqrt[3]4}=\frac 12\Bigg(1+\sqrt{\frac 5{\sqrt[3]4}-1}\Bigg)$$ and I put two and two together.

Of course, nobody just notices these things (except maybe Ramanujan). I was simply doing what I described earlier about deriving these backwards and merely experimenting and playing around with numbers for the fun of it. But I really want to know why these outputs do come out so nicely, and the essence of it all.
Any thoughts?
Thank you in advance.
 [/INST] Well, let's do this step by step:


*

*Write:
$$\sqrt[3]{4}=\sqrt[3]{2^2}=2^\frac{2}{3}\tag1$$

*Write:
$$\frac{5}{2^\frac{2}{3}}=\frac{5}{2^\frac{2}{3}}\cdot\frac{\sqrt[3]{2}}{\sqrt[3]{2}}=\frac{5\sqrt[3]{2}}{2}\tag2$$

*Write:
$$\left(3-\sqrt[3]{2}\right)\left(\sqrt[3]{2}-1\right)=-3+3\sqrt[3]{2}-\left(-\sqrt[3]{2}\right)-\sqrt[3]{2}\sqrt[3]{2}=-3+3\sqrt[3]{2}+\sqrt[3]{2}-2^\frac{2}{3}=$$
$$-3+4\sqrt{3}{2}-2^\frac{2}{3}=1+2\sqrt[3]{2}-2^\frac{2}{3}-4+2\sqrt[3]{2}=$$
$$1+2\sqrt[3]{2}-\left(\sqrt[3]{2}\right)^2-2\left(\sqrt[3]{2}\right)^3+\left(\sqrt[3]{2}\right)^4=\left(1+\sqrt[3]{2}-2^\frac{2}{3}\right)^2\tag3$$

*Write:
$$\frac{5\sqrt[3]{2}}{2}-1=\frac{5\sqrt[3]{2}}{2}-\frac{2}{3}=\frac{5\sqrt[3]{2}-2}{2}\tag4$$

*Write:
$$5\sqrt[3]{2}-2=2+4\sqrt[3]{2}-4+\sqrt[3]{2}=\frac{4+8\sqrt[3]{2}-8+2\sqrt[3]{2}}{2}=$$
$$\frac{4+8\sqrt[3]{2}-4\left(\sqrt[3]{2}\right)^3+\left(\sqrt[3]{2}\right)^4}{2}=\frac{\left(2+2\sqrt[3]{2}-2^\frac{2}{3}\right)^2}{2}\tag5$$

I think you can finish.

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