Question

How do you solve x(bc)+y(ac)+z(ab)=abc for a?

Answer 1

well, first distribute the factors, so \(\bf x(bc)+y(ac)+z(ab)=abc \implies xbc+yac+zac=abc
\\ \quad \\
y{\color{brown}{ a}}c+z{\color{brown}{ a}}c-{\color{brown}{ a}}bc=-xbc\implies {\color{brown}{ a}}(yc+zc-bc)=-xbc\)

and I'm sure you can see what "a" is there already