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.
 A: Unfortunately, you do have to expand the brackets.
Fortunately, it's not so messy after all:
$$(x+2)(x-3)(x-1)+4=0$$
so 
$$x^3-2x^2-5x+10=0$$
and factoring our $x-2$, 
$$(x-2)(x^2-5)=0$$
from which we can readily read the answer as $x=2,\pm\sqrt{5}$
A: 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.
A: It is good that you managed to factor!
Here is one way to do it:
$$(x+a)(x+b)(x+c)+(b+c)(c+a)(a+b)=\\
x^3+(a+b+c)x^2+(ab+bc+ca)x+abc+\\
2abc+a^2b+ab^2+a^2c+ac^2+b^2c+bc^2=\\
x^2(x+a+b+c)+(ab+bc+ca)x+\\
ab(a+b+c)+ac(a+b+c)+bc(a+b+c)=\\
x^2(x+a+b+c)+(ab+bc+ca)x+\\
(a+b+c)(ab+bc+ca)=\\
x^2(x+a+b+c)+(ab+bc+ca)(x+a+b+c)=\\
(x+a+b+c)(x^2+ab+bc+ca).$$
Now, write the given equation in this form:
$$(x+2)(x-3)(x-1)+4=0 \iff \\
(x+2)(x-3)(x-1)+(2-3)(-3-1)(-1+2)=0 \iff \\
(x+2-3-1)(x^2+2(-3)+(-3)(-1)+(-1)2)=0 \iff \\
(x-2)(x^2-5)=0 \Rightarrow x_1=2, x_{2,3}=\pm \sqrt{5}.$$
A: Probably not more efficient than straightforwardly factoring out, but it works as well:
$$\begin{eqnarray}\frac{(x+2)(x−3)(x−1)+4}{x-2} & = & \frac{(x-2+4)(x−3)(x−1)}{x-2}+\frac{4}{x-2} \\
& = & (1+\frac{4}{x-2})(x−3)(x−1)+\frac{4}{x-2} \\
& = & (x−3)(x−1) + \frac{4(x−3)(x−2+1)}{x-2}+\frac{4}{x-2}\\
& = & (x−3)(x−1) + 4(x−3)(1+\frac{1}{x-2})+\frac{4}{x-2}\\
& = & (x−3)(x−1) + 4(x−3) + \frac{4(x−2-1)}{x-2}+\frac{4}{x-2}\\
& = & (x−3)(x−1) + 4(x−3) + 4\\
& = & (x−3)(x+3) + 4\\
& = & x^2 - 5 \; .
\end{eqnarray}$$
