<s>[INST] Expressing in terms of symmetric polynomials. How to express $$a^7+b^7+c^7$$  in terms of symmetric polynomials ${\sigma}_{1}=a+b+c$, ${\sigma}_{2}=ab+bc+ca$ and ${\sigma}_{3}=abc$ ?
 [/INST] Hint: Expand $(a+b+c)^7=\sigma_1^7$ with Newton to find the terms you need to subtract from it to get $a^7+b^7+c^7.$ I think it's not too high of a power to do that, since its expansion has $\frac{8\cdot9}{2}=36$ terms, arguably not too many, but after all one can also compute $(a+b+c)^6$ by squaring $(a+b+c)^3$ and then multiply by $a+b+c=\sigma_1$.
For example, for the square we have
$$\require\color(a+b+c)^2=\sigma_1^2=a^2+b^2+c^2+2ab+2ac+2bc =a^2+b^2+c^2+2(ab+bc+ca)  \iff \ \color\red{a^2+b^2+c^2=\sigma_1^2-2\sigma_2}$$
and for the cube
$$\require\color (a+b+c)^3=\sigma_1^3=a^3+b^3+c^3+3ab^2+3ac^2+3ba^2+3bc^2+3ca^2+3cb^2+6abc=\\ a^3+b^3+c^3+3\left(ab(a+b)+bc(b+c)+ca(c+a)\right)+6abc=\\ a^3+b^3+c^3+3(a+b+c)(ab+bc+ca)-9abc+6abc \iff \\ \color\red{a^3+b^3+c^3 = \sigma_1^3-3\sigma_1\sigma_2+3\sigma_3}.$$
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