Given $f(x) = \frac{x^3}{3} - \sqrt{x^2+1}$, proof that $f(\alpha) = \frac{\alpha^4 - 3}{3\alpha}$ I evaluated the function on $\alpha$ and removed the radical by multiplying by $\sqrt{\alpha^2+1}$ as follows:
$$ \frac{\alpha}{3} - \frac{\alpha^2 + 1}{\sqrt{\alpha^2+1}} $$
After some algebraic manipulations, I got to this step:
$$ \frac{\alpha^4 \sqrt{1 + 1/\alpha^2}}{3\alpha\sqrt{1 + 1/\alpha^2}} - \alpha(1 + 1/\alpha^2)^{-1/2} $$
Does anyone knows if I am on the good track? or is there any hints that could help me? Thanks

$$\frac{x^3}{3}-\sqrt{x^2+1}=\frac{x^4-3}{3 x}$$
$$\frac{x^3}{3}-\frac{x^4-3}{3 x}=\sqrt{x^2+1}$$
$$\frac{1}{x^2}=x^2+1$$
Substitute $x^2=w$
$$\frac{1}{w}=w+1\to w^2+w-1=0\to w=\frac{1}{2}\left(\sqrt{5}-1\right)$$
the root $w_2=\frac{1}{2}\left(-\sqrt{5}-1\right)$ is discarded because is negative. The solution is:
$$x=\sqrt{\frac{1}{2}\left(\sqrt{5}-1\right)}$$
