Calculate $a^3+b^3+c^3+d^3$ for the real roots of $x^4+2x^3-3x^2-3x+2$ $a,b,c,d \in \mathbb{R}$ are the real roots of $x^4+2x^3-3x^2-3x+2$. 
Calculate $a^3+b^3+c^3+d^3$.
With approximation i found out, that $a^3+b^3+c^3+d^3 = -17$, but how can I proof that without calculating the roots exactly?
Cheers

Express $a^3 + b^3 + c^3 + d^3$ in terms of elementary symmetric polynomials and use Vieta's formulas.
Also, here is an alternative approach for the fans of linear algebra. Consider a matrix
$$
A = \left(\matrix{0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ -2 & 3 & 3 & -2}\right).
$$
A trivial check shows that its characteristic polynomial is equal to
$$
\det (A - \lambda I) = \lambda^4 + 2\lambda^3 - 3\lambda^2 - 3\lambda + 2.
$$
So $a,b,c,d$ are the characteristic roots of $A$. Then their cubes are characteristic roots of $A^3$ (as easily follows from, say, the existence of a Jordan normal form). Then
$$
a^3 + b^3 + c^3 + d^3 = \operatorname{Tr} A^3 = -17.
$$
