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\frac{8\left(2+\sqrt{6}\right)}{\left(2-\sqrt{6}\right)\left(2+\sqrt{6}\right)}
Rationalize the denominator of \frac{8}{2-\sqrt{6}} by multiplying numerator and denominator by 2+\sqrt{6}.
\frac{8\left(2+\sqrt{6}\right)}{2^{2}-\left(\sqrt{6}\right)^{2}}
Consider \left(2-\sqrt{6}\right)\left(2+\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}.
\frac{8\left(2+\sqrt{6}\right)}{4-6}
Square 2. Square \sqrt{6}.
\frac{8\left(2+\sqrt{6}\right)}{-2}
Subtract 6 from 4 to get -2.
-4\left(2+\sqrt{6}\right)
Divide 8\left(2+\sqrt{6}\right) by -2 to get -4\left(2+\sqrt{6}\right).
-8-4\sqrt{6}
Use the distributive property to multiply -4 by 2+\sqrt{6}.