How to retrieve the expression of $a^3+b^3+c^3$ in terms of symmetric polynomials? I recently had to find without any resources the expression of $a^3+b^3+c^3$ in terms of $a+b+c$, $ab+ac+bc$ and $abc$.
Although it's easy to see that $a^2+b^2+c^2=(a+b+c)^2-2(ab+ac+bc)$, I couldn't come up myself with $a^3+b^3+c^3=(a+b+c)^3-3(a+b+c)(ab+ac+bc)+3abc$.
Can someone explain to me how to retrieve this formula with only pen and paper, or how to memorize it for good ?
 A: In terms of the products of the three polynomials, the only possible way of getting an $a^3$ term is from $$(a+b+c)^3=a^3+b^3+c^3+3\sum a^2b+6abc$$
Where the sum is of all the expressions of the same kind. To obtain $a^2b$ from the symmetric polynomials, you need $a\cdot ab$ which comes from $$(a+b+c)\cdot(ab+bc+ca)=\sum a^2b+3abc$$ and substituting the expression for $\sum a^2b$ into the first equation gives $$(a+b+c)^3=a^3+b^3+c^3+3(a+b+c)(ab+bc+ca)-3abc$$ Which is what you want. Note that we started by eliminating the term in which $a$ appeared to the highest power - $a^3$ - and then went to $a^2$ and then we were done.
For higher degrees and more variables there is an entirely systematic process which works as a generalisation of this method.
A: Let $E$ be the expression equal to $a^3+b^3+c^3$ in terms of symmetric polynomials. Since the reduced expression has no mixed terms, you must have $E = (a+b+c)^3 + E'$, where
$$ E' = a^3+b^3+c^3 - (a+b+c)^3 = -3(a^2b +ab^2 +a^2 c +2abc +b^2c +ac^2 +bc^2) $$
To get all of the terms like $a^2b$, we must have $(a+b+c)(a^2+b^2+c^2)$ somewhere in $E'$; after calculating this, the final answer should be clear.
