Factoring a trinomial with three different variables (grade 10)
Solution 1:
I started by seeing that $X=x^3y^3$ comes in this equation as follows: $$X^2- 4 X z^2-12z^4.$$ This is a quadratic equation and thus simple to solve, the reason to introduce this $X$. Solving this for $X$, we get the following roots: \begin{align*} X_\pm=(2\pm 4) z^2 \end{align*} so that $$X^2- 4 X z^2-12z^4 = (X-X_+)(X-X_-)= (X-6z^2)(X+2z^2)$$ or $$x^6y^6- 4 x^3y^3 z^2-12z^4 = (x^3y^3-6z^2)(x^3y^3+2z^2).$$