<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] 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
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