vieta's formula is considered as basic knowledge for an a-level student. interestingly, most of the relevant questions are actually asking for newton's sum. as i checked among most a-level and equivalent curricula, it's never seriously mentioned.

basically, if polynomial P(x)=\sum\limits_{i=0}^na_{n-i}x^i has roots x_i, i\in[n], then the **newton's sum** s_k is defined as s_k=\sum\limits_{i=1}^nx_i^k . **newton-girard identity** states that:

the result is obvious for k\ge n as a_s=0, \forall s < 0 , as

\sum_{i=1}^nx_i^{k-n}P(x_i) = 0the expansion of the l.h.s. is equivalent to that of newton-girard.

among all proofs for k < n, the differentiation method is the most straight-forwarded:

notice that if x_i, i\in[n] are the roots, P(x)=a_0\prod\limits_{i=1}^n(x-x_i). now that we differentiate both side of the identity

a_0\prod_{i=1}^n(x-x_i)=\sum_{i=0}^na_{n-i}x^ithe l.h.s. yields

\frac{\mathrm dP(x)}{\mathrm dx}=\sum_{i=1}^n\frac{P(x)}{x-x_i}and r.h.s. yields

\frac{\mathrm dP(x)}{\mathrm dx}=\sum_{i=1}^n ia_{n-i}x^{i-1}we can obtain the identity by comparing the coefs of like terms.