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.

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:
$$(x+2)(x-3)(x-1)+4=0 \iff \\
(x+2)(x-3)(x-1)+(2-3)(-3-1)(-1+2)=0 \iff \\
(x+2-3-1)(x^2+2(-3)+(-3)(-1)+(-1)2)=0 \iff \\
(x-2)(x^2-5)=0 \Rightarrow x_1=2, x_{2,3}=\pm \sqrt{5}.$$
