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
 A: 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
$$
a^3 + b^3 + c^3 + d^3 = \operatorname{Tr} A^3 = -17.
$$
A: As biquadratic equation has exactly $4$ roots, all of $a,b,c,d$ are real
Clearly, $abcd\ne0$
and $\displaystyle a^4+2a^3-3a^2-3a+2=0\implies a^3+2a^2-3a-3=-\frac2a$
Similarly, for $b,c,d$
Summing we get $\sum a^3+2\sum a^2-3\sum a-3\sum 1=-2\sum \frac 1a$
Now, we need 
$\sum a$
$\displaystyle\sum\frac1a=\frac{\sum abc}{abcd}$
$\displaystyle\sum a^2=(\sum a)^2-2\sum ab$
Vieta's formulas is calling for usage
A: thanks for your help.
$P(x) = x^4 + 2x^3 -3x^2 -3x + 2:$
$\sigma_1 = -2$
$\sigma_2 = -3$
$\sigma_3 = 3$
$\sigma_4 = 2$
$a^3 + b^3 + c^3 + d^3 = \sigma_1^3 - 3*\sigma_1 * \sigma_2 + 3 * \sigma_3$
We get: $a^3 + b^3 + c^3 + d^3 = (-2)^3 - 3 * (-2) * (-3) + 3 * 3 = -17$
