Consider $\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)$. Multiplication can be transformed into difference of squares using the rule: $\left(a-b\right)\left(a+b\right)=a^{2}-b^{2}$.
Consider $\left(\sqrt{11}+\sqrt{7}\right)\left(\sqrt{11}-\sqrt{7}\right)$. Multiplication can be transformed into difference of squares using the rule: $\left(a-b\right)\left(a+b\right)=a^{2}-b^{2}$.
Since $\frac{2\sqrt{6}\left(\sqrt{7}-\sqrt{5}\right)}{2}$ and $\frac{2\left(\sqrt{11}-\sqrt{7}\right)}{2}$ have the same denominator, add them by adding their numerators.
