I would like to have some help on this problem:
Given the following system: $$ \begin{aligned} a + b + c &= 6\\ a^2 + b^2 + c^2 &= 14\\ a^3 + b^3 + c^3 &= 36 \end{aligned} $$
Without guessing or using computer can we solve this system ?
Thanks for your help
Note that the left-hand side of each of the three equations is a symmetric polynomial in $a, b$ and $c$. These can be written in terms of the elementary symmetric polynomials in $a, b$ and $c$, which are \begin{align*} u &= a+ b + c \\ v &= ab + bc + ca \\ w &= abc. \end{align*} This way, we obtain the following system \begin{align*} u &= 6 \\ u^2 - 2v &= 14 \\ u^3 + 3w - 3uv &= 36 \end{align*} which has a unique solution, namely $u = 6, v = 11$ and $w = 6$. So, we have to solve the system \begin{align*} a+ b + c &= 6\\ ab + bc + ca &= 11\\ abc &= 6. \end{align*} By Vieta's formulas, $a, b$ and $c$ are the roots of the polynomial equation $$ x^3 - 6x^2 + 11x - 6 = 0. $$ Since $x = 1, x = 2$ and $x = 3$ are the three roots of this equation, we can conclude that the only solution to the given system is $(1, 2, 3)$, up to permutation.