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

$$({x^2} - 1)({x^2} - 3x - 1) = 0$$

Let the root of $${x^2} - 3x - 1 = 0$$ be $$\alpha$$ and $$\beta$$ and other two roots of given equation are 1 and $$-$$1

So sum of cubes of roots

$$ = {1^3} + {( - 1)^3} + {\alpha ^3} + {\beta ^3}$$

$$ = {(\alpha + \beta )^3} - 3\alpha \beta (\alpha + \beta )$$

$$ = {(3)^3} - 3( - 1)(3)$$

$$ = 36$$