Consider $\left(\sqrt{4}-\sqrt{6}\right)\left(\sqrt{6}+\sqrt{4}\right)$. Multiplication can be transformed into difference of squares using the rule: $\left(a-b\right)\left(a+b\right)=a^{2}-b^{2}$.
Consider $\left(9-3\sqrt{6}\right)\left(9+3\sqrt{6}\right)$. Multiplication can be transformed into difference of squares using the rule: $\left(a-b\right)\left(a+b\right)=a^{2}-b^{2}$.
