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.

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.

