problem to Brilliant ! 5

@Mehul Chaturvedi – For simplicity lets just assume that there are 3 roots(although this can be generalised easily). Let $S_n$ denote the sum of the $n^{\text{th}}$ powers of the roots, i.e. $S_n=\alpha^n+\beta^n+\gamma^n$ let $\alpha, \beta, \gamma$ be the roots of $f(x)=ax^3+b^2+cx+d=0$ Since $\alpha, \beta, \gamma$ are the roots of $f(x)=0$ , $f(\alpha)=a\alpha^3+b\alpha^2+c\alpha+d=0$ $f(\beta)=a\beta^3+b\beta^2+c\beta+d=0$ $f(\gamma)=a\gamma^3+b\gamma^2+c\gamma+d=0$ Adding all the eqns, we get $aS_3+bS_2+cS_1+3d=0$ What I did above can also be stated as plugging the roots in succession and then adding.

Well, if we know $S_2$ and $S_1$ we can easily calculate $S_3$ . But we could calulate $S_3$ even without that, so how is that useful? In the next step we draw some useful results and try to generalize a bit. We started with $f(x)=ax^3+b^2+cx+d=0$ what if multiply both sides by $x^{n-3}$ ? Lets just try that $ax^n+b^{n-1}+cx^{n-2}+dx^{n-3}=0$ Since we just multiplied the equation by a number, this still hasn't affected anything(although we have introduced more roots), this means that $\alpha, \beta ,\gamma$ are still its roots.

Now we plug-in the roots in succession and add.

$aS_{n}+bS_{n-1}+cS_{n-2}+dS_{n-3}=0$

This is a very useful result and it can help us in calculating $S_{n}$ if we know $S_{n-1}$ , $S_{n-2}$ , $S_{n-3}$ (and also the coefficients of the polynomial).

Hope this helped.