Express $a^2b+a^2c+b^2a+b^2c+c^2a+c^2b$ in terms of elementary symmetric polynomials.

Question

Express $a^2b+a^2c+b^2a+b^2c+c^2a+c^2b$ in terms of elementary symmetric polynomials.

I started with expanding $(a + b + c)^3$ and got to

$a^2b+a^2c+b^2a+b^2c+c^2a+c^2b = \frac{1}{3}(a + b + c)^3 -a^3 -b^3 - c^3 -2abc$

I see that the first and last terms on the right hand side are expressions of $S_1$ and $S_3$, but I am having trouble with the $-a^3 -b^3 - c^3$ terms.

Answer 1

$$\begin{align}S_1S_2&=(a+b+c)(ab+ac+bc)\\ &=a^2b+a^2c+abc+ab^2+abc+b^2c+abc+ac^2+bc^2\\ &=a^2b+a^2c+ab^2+b^2c+ac^2+bc^2+3S_3\end{align}$$

Answer 2

$a^2b+a^2c+b^2a+b^2c+c^2a+c^2b$$ $$=$$ $$(a+b+c)(ab+bc+ca)-3abc$

Answer 3

Note,

$$\begin{array} && a^2b+a^2c+b^2a+b^2c+c^2a+c^2b \\ & =ab(a+b)+bc(b+c) + ca(c+a) \\ & =ab(a+b+c-c)+bc(b+c+a-a) + ca(c+a+b-b) \\ & =ab(a+b+c)-abc+bc(b+c+a)-bca + ca(c+a+b) - cab\\ & =(ab+bc+ca)(a+b+c)-3abc \end{array}$$