Trying to understand this passage where this:
$$\left(\frac{11}{12}\beta-I\right)\left[\left(\frac{2}{3}\beta-I\right)^2 -\frac{1}{8}\beta^2-\frac{1}{4}\beta\left(\frac{2}{3}\beta-I\right)\right]=0$$
gets factored into this:
$$\left(\frac{1}{6}\beta-I\right) \left(\frac{11}{12}\beta-I\right) \left(\frac{11}{12}\beta-I\right)=0$$
I is a real variable and $\beta$ is a constant. What method is used to achieve this factorization?
We have by expanding and simplifying
$$\left(\frac{11}{12}\beta-I\right)\left[\left(\frac{2}{3}\beta-I\right)^2 -\frac{1}{8}\beta^2-\frac{1}{4}\beta\left(\frac{2}{3}\beta-I\right)\right]$$ $$=\left(\frac{11}{12}\beta-I\right)\left[\frac{4}{9}\beta^2-\frac{4}{3}\beta I+I^2-\frac{1}{8}\beta^2-\frac{1}{6}\beta^2+\frac{1}{4}\beta I\right]$$ $$=\left(\frac{11}{12}\beta-I\right)\left[\frac{11}{72}\beta^2-\frac{13}{12}\beta I+I^2\right]$$ $$=\left(\frac{11}{12}\beta-I\right)\left[\frac{11}{72}\beta^2-\frac{1}{6}\beta I-\frac{11}{12}\beta I +I^2\right]$$ $$\left(\frac{11}{12}\beta-I\right)\left[\frac{1}{6}\beta\left(\frac{11}{12}\beta-I\right)-I\left(\frac{11}{12}\beta-I\right)\right]$$ $$=\left(\frac{1}{6}\beta-I\right)\left(\frac{11}{12}\beta-I\right)\left(\frac{11}{12}\beta-I\right)$$