Does this radical ${a-1\over 2}\left({\sqrt[3]a+1\over \sqrt[3]a-1}-1\right)=...$ hold?

Question

Consider the nested radical $(1)$

$${a-1\over 2}\left({\sqrt[3]a+1\over \sqrt[3]a-1}-1\right)=2\sqrt[3]a+\left[(a^2-7a+1)+(6a-3)\sqrt[3]a+(6-3a)\sqrt[3]{a^2}\right]^{1/3}\tag1$$

An attempt:

$x=\sqrt[3]a$

$$\left({a-1\over 2}\left({x+1\over x-1}-1\right)-2x\right)^3=...\tag2$$

$(2)$ on the LHS becomes

$$\left({a-1\over x-1}-2x\right)^3=\left({a-1\over x-1}\right)^3-6x\left({a-1\over x-1}\right)^2+12x^2\left({a-1\over x-1}\right)-8x^3$$

We can't and don't how to simplify it any further.

Can anyone help us to prove that $(1)$ does hold?

Answer 1

Using your substitution, you will find that the LHS simplifies to $x^2-x+1$, while the RHS simplifies to the cube root of a sextic in $x$, which you can straightforwardly check to be equal to $(x^2-x+1)^3$.

Answer 2

$y^3-1=(y-1)(y^2+y+1)$ ... let $y=\sqrt[3] a$. So I guess your LHS is \begin{eqnarray*} \frac{a-1}{2} \left( \frac{\sqrt[3] a+1 }{\sqrt[3] a-1}-1 \right) = (\sqrt[3] a)^2+\sqrt[3] a+1 \end{eqnarray*} Now subtract $2 \sqrt[3] a$ from both sides and then cube both sides ... you will get the second term of the RHS ... but the algebra is pretty heavy.