<s>[INST] Proving $\sqrt[3]{\sqrt[3]{2}-1}=\sqrt[3]{\frac19}-\sqrt[3]{\frac29}+\sqrt[3]{\frac49}$ I found the following two relational expressions in a book without any additional information:
$$\sqrt{\sqrt[3]{5}-\sqrt[3]{4}}=\frac13(\sqrt[3]{2}+\sqrt[3]{20}-\sqrt[3]{25})$$
$$\sqrt[3]{\sqrt[3]{2}-1}=\sqrt[3]{\frac19}-\sqrt[3]{\frac29}+\sqrt[3]{\frac49}$$
Wolfram tells these are true, but I can't prove at all. Can anyone help?
 [/INST] 
For the identity in the title, use the identity $x^3+1=(x+1)(x^2-x+1)$.

Let $x=\sqrt[3]{2}$, then $(x^2-x+1)(x+1)=x^3+1=3$ hence the RHS is $\sqrt[3]{3}/(x+1)$ and the LHS divided by the RHS is the cube root of one third of $(x-1)(x+1)^3=(x^3-2)(x+2)+3$. Since $x^3=2$, this is $3$ and the result follows.

For the other identity:

Let $x=\sqrt[3]{5/4}$, then the identity to prove is $3\sqrt{x-1}=1+2x-2x^2$. Squaring both sides, this is equivalent to $1+4(x-x^2)+4(x-x^2)^2-9(x-1)=(5-4x^3)(2-x)$ being zero. Since $4x^3=5$, the result follows.
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