Q.Find $\alpha^2+\beta^2+\gamma^2$, where $\alpha,\beta,\gamma$ are the roots of $x^3-6x^2+11x-6=0$, without solving the equation.

If $\alpha,\beta,\gamma$ are the roots of the monic cubic $x^3+px^2+qx+r=0$, then $f(x)\equiv(x-\alpha)(x-\beta)(x-\gamma)$. Expanding the right side and matching coefficients with $x^3+px^2+qx+r$ gives

For the biquadratic $x^4+px^3+qx^2+rx+s=0$ with roots $\alpha,\beta,\gamma,\delta$, expanding $(x-\alpha)(x-\beta)(x-\gamma)(x-\delta)$ in the same way gives

In general, for the monic degree-$n$ equation $x^n+p_1x^{n-1}+p_2x^{n-2}+\cdots+p_n=0$ with roots $\alpha_1,\ldots,\alpha_n$,

$S_k=\sum(\text{products of the roots taken }k\text{ at a time})=(-1)^kp_k,\qquad k=1,\ldots,n.$ …