<s>[INST] 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
 [/INST] 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: $a^3 + b^3 + c^3 + d^3 = (-2)^3 - 3 * (-2) * (-3) + 3 * 3 = -17$
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