Let $a$ be the real root of the equation $x^3+x+1=0$

Maybe there is a trick to it, but good ol' factoring can work here.

It is easy to check (using the identity $a^3=-a-1$) that $$ (3a^2-2a+2)(3a^2+2a) = -5a-7a^2 $$ Let $\sqrt[3]{-5a-7a^2}+a^2 = y$. Given that $a\in \mathbb{R}$, we know that also $y\in \mathbb{R}$, and we must solve for it. $$y-a^2 = \sqrt[3]{-5a-7a^2} \\ (y-a^2)^3 = -5a-7a^2 \\ y^3 - 3y^2a^2 + 3ya^4-a^6 = -5a-7a^2 \\ y^3-3y^2a^2+3ya(-a-1)-(-a-1)^2=-5a-7a^2 \\ y^3-3y^2a^2-3ya^2-3ya+6a^2+3a-1=0$$ So many threes, only the $6$ stands out. But if we use $6=3+3$: $$ (y^3-1)+(3a^2-3y^2a^2)+(3a^2+3a-3ya^2-3ya)=0 \\ (y^3-1)-3a^2(y^2-1)-(3a^2+3a)(y-1)=0$$ As one can see, it is possible to factor out $(y-1)$, thereby concluding that $y=1$ is a solution.

The expansion of what is under square root gives

$$A=9a^4+6a^3-6a^3-4a^2+6a^2+4a=$$ $$a (9a^3+2a+4)=a (9 (-a-1)+2a+4)=$$ $$-a (7a+5)=$$

this must be equal to $$B=(1-a^2)^3=1-3a^2+3a^4-a^6=$$ $$$$

the difference is $$B-A=1+4a^2+5a+3a^4-(a+1)^2=$$ $$3a^2+3a+3a^4=3a (a^3+a+1)=0$$

It is correct.

The real root $a$ of $x^3+x+1$ belongs to the interval $\left(-1,-\frac{1}{2}\right)$ and in order to check that $$ \sqrt[3]{(3a^2-2a+2)(3a^2+2a)}=1-a^2 \tag{1}$$ holds it is enough to check that $$ (3a^2-2a+2)(3a^2+2a)=1-3a^2+3a^4+a^6 \tag{2}$$ holds, or that $$ (1+a+a^3)(-1+5a+a^3) = 0 \tag{3} $$ holds, but that is trivial.