The sum of the cubes of all the roots of the equation x 4 - 3 x 3 - 2 x 2 + 3 x + 1 = 0 is _____.

$x^4-3x^3-2x^2+3x+1=0$

Dividing by $x^2$ as $x\ne 0$

$x^2-3x-2+\frac{3}{x}+\frac{1}{x^2}=0$

$x^2+\frac{1}{x^2}-2+2-3\left(x-\frac{1}{x}\right)-2=0$

${\left(x-\frac{1}{x}\right)}^2-3\left(x-\frac{1}{x}\right)=0$

$\begin{array}{cc}x-\frac{1}{x}=0,& x-\frac{1}{x}=3\end{array}$

$\begin{array}{cc}{x}^2-1=0,& x^2-3x-1=0\end{array}$

$x=\pm 1$

For $x^2-3x-1=0$ let the roots are $\gamma \& \delta$

$\begin{array}{cc}\alpha =1,\beta =-1, \gamma +\delta =3,& \gamma \delta =-1\end{array}$

${\alpha }^3+{\beta }^3+{\gamma }^3+{\delta }^3$

$=1-1+\left(\gamma +\delta \right)\left({\left(\gamma +\delta \right)}^2-3\gamma \delta \right)$

$=3\left(9-3\left(-1\right)\right)$$=36$