<s>[INST] Factoring $(x+a)(x+b)(x+c)+(b+c)(c+a)(a+b)$ and use the result to solve an equation I managed to prove that $(x+a+b+c)$ is a factor of
$$(x+a)(x+b)(x+c)+(b+c)(c+a)(a+b)$$
Then I was asked to use the result to solve
$$(x+2)(x-3)(x-1)+4=0$$
I know by comparison, $a=2, b=-3, c=-1$, and thus $(x-2)$ is a factor, but I can't really figure out how to solve the equation without expanding the brackets.
 [/INST] Since the $x^2$ coefficient in your cubic is $a+b+c$, the quadratic factor is of the form $x^2+k$, with the roots being $\pm\sqrt{-k}$. The $x=0$ case gives $$k=\frac{abc+(b+c)(c+a)(a+b)}{a+b+c}=ab+bc+ca.$$In your case $k=-5$.
Simplifying $k$ as above looks like it requires tedious algebra, but things aren't as bad as they seem. It's the ratio of two fully symmetric polynomials in $a,\,b,\,c$, one of degree $3$, the other $1$. This doesn't prove on its own that $k$ is a polynomial; but if it is, it must be fully symmetric and of degree $2$, and hence proportional to $ab+bc+ca$. The case $a=b=c$ gives $k=\frac{9a^3}{3a}=3a^2$, so it'll have to be $ab+bc+ca$ itself. So it makes sense to double-check whether $(ab+bc+ca)(a+b+c)=abc+(b+c)(c+a)(a+b)$. But of course it does, because both sides are fully symmetric cubic functions, so they have a fixed ratio. Again, the case $a=b=c$ strengthens this to equality.
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