The sum of the cubes of all the roots of the equation $x^4-3x^3-2x^2+3x+1=10$ is _________

$x^4-3x^3-2x^2+3x+1=10$

$\mathrm{x}=0\text{ is not the root of this equation so divide it by }{\mathrm{x}}^2$

$\begin{array}{l}{x}^2-3x-2+\frac{3}{x}+\frac{1}{x^2}=0\\ x^2+\frac{1}{x^2}-2+2-3\left(x-\frac{1}{x}\right)-2=0\\ {\left(x-\frac{1}{x}\right)}^2-3\left(x-\frac{1}{x}\right)=0\\ x-\frac{1}{x}=0,x-\frac{1}{x}=3\\ x^2-1=0,x^2-3x-1=0\\ \mathrm{x}=\pm 1,\gamma +\delta =3\\ \alpha =1,\beta =-1\gamma \delta =-1\\ {\alpha }^3+{\beta }^3+{\gamma }^3+{\delta }^3\\ =1-1+(\gamma +\delta )\left((\gamma +\delta {)}^2-3\gamma \delta \right)\\ 0+3(9-3(-1\left)\right)\\ =3\left(12\right)=36\end{array}$