The main question is :
$$\begin{align} (b+c)(y+z)-ax &= b-c \tag{1} \\ (c+a)(z+x)-by &= c-a \tag{2} \\ (a+b)(x+y)-cz &= a-b \tag{3}\\ \end{align}$$
Solve for $x,y,z$ if $a+b+c\ne0$
My method :
Opening all brackets, we get,
$$by+bz+cy+cz-ax=b-c$$ $$cz+cx+az+ax-by=c-a$$ $$ax+ay+bx+by-cz=a-b$$
Adding, $$(a+b+c)(x+y+z)=0$$ Thus, $$(x+y+z)=0$$
Also, $$by+bz+cy+cz-b+c=by$$ Thus, $$b(y+z-1)+c(y+z+1)=ax$$ Similarly, $$c(z+x-1)+a(z+x+1)=by$$ $$a(x+y-1)+b(x+y+1)=cz$$
I can't go any further, nothing is clicking. Please help me.
You're almost there; nice work!
Plug $x+y+z = 0$, in the form $(y+z) = -x$, into your first equation to get $(-x)(a+b+c) = b-c,$ so $$x = \frac{c-b}{a+b+c}.$$
Do similar things to the other two equations.