Polynomial $x^3-2x^2-3x-4=0$

By long division, the remainder of $x^6$, when divided by $x^3-2x^2-3x-4$, is $77x^2+100x+96$. So we know that $$ \alpha^6=77\alpha^2+100\alpha+96 $$ and the same with other roots. Therefore you are looking at the sum $$ \begin{aligned} S&=\frac{77(\alpha^2-\beta^2)+100(\alpha-\beta)}{\alpha-\beta}+\text{cyclic}\\ &=77(\alpha+\beta)+100+\text{cyclic}\\ &=154(\alpha+\beta+\gamma)+300\\ &=608 \end{aligned} $$ by the (Vieta) relations you have.


If $\alpha,\beta,\gamma$ satisfy $x^6=ax^2+bx+c$, then the first term is $a(\alpha+\beta)+b$, and we have similar expressions for the others, so the desired sum is $2a(\alpha+\beta+\gamma)+3b=4a+3b$. To find $a$ and $b$, square $x^3$ and reduce modulo the given polynomial.

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Polynomials

Symmetric Polynomials

Roots

Irreducible Polynomials

Cubic Equations

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