Value of $(3\beta^2-4\beta)^{\frac{1}{3}}+(3\beta^2+4\beta+2)^{\frac{1}{3}}$ if $\beta$ is the root of $x^3-x-1=0$

The trick is that if we reduce $(x + a)^3$ modulo $x^3 - x - 1$, we'll get $3 a x^2 + (3 a^2 + 1) x + (a^3 + 1)$. Therefore, if $\beta$ is a root of $x^3 - x - 1$, $$(1 + \beta)^3 = 3 \beta^2 + 4 \beta + 2, \\ (1 - \beta)^3 = 3 \beta^2 - 4 \beta.$$ Then we need to define the cube root in such a way that $(z^3)^{1/3} = z$ for $z = 1 \pm \beta$.

I assume that $\beta$ is the real root of $x^3-x-1=0$ so that the cube roots are well defined.

Let $\mu = 3\beta^2-4\beta$ and $\nu = 3\beta^2+4\beta+2$. We seek $\tau = \mu^{\frac{1}{3}}+\nu^{\frac{1}{3}}$. As in this question, we have $$ \tau^3 = \mu+\nu+3(\mu\nu)^{\frac{1}{3}}\tau $$ Hoping that $\mu\nu = \beta-\beta^2$ is a cube in $\mathbb Q(\beta)$, we find after some work that $\beta-\beta^2=(1-\beta^2)^3$. Therefore, $$ \tau^3 = \mu+\nu+3(\mu\nu)^{\frac{1}{3}} = 6\beta^2+2+3(1-\beta^2)\tau $$ Now $\tau=2$ is a root. The other roots are complex because the discriminant of the quotient quadratic is $-12\beta^2$.

Therefore, $\tau=2$ is the only real root and the answer is $2$.