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\frac{8\sqrt{2}\left(\sqrt{6}-\sqrt{2}\right)}{\left(\sqrt{6}+\sqrt{2}\right)\left(\sqrt{6}-\sqrt{2}\right)}
Rationalize the denominator of \frac{8\sqrt{2}}{\sqrt{6}+\sqrt{2}} by multiplying numerator and denominator by \sqrt{6}-\sqrt{2}.
\frac{8\sqrt{2}\left(\sqrt{6}-\sqrt{2}\right)}{\left(\sqrt{6}\right)^{2}-\left(\sqrt{2}\right)^{2}}
Consider \left(\sqrt{6}+\sqrt{2}\right)\left(\sqrt{6}-\sqrt{2}\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\sqrt{2}\left(\sqrt{6}-\sqrt{2}\right)}{6-2}
Square \sqrt{6}. Square \sqrt{2}.
\frac{8\sqrt{2}\left(\sqrt{6}-\sqrt{2}\right)}{4}
Subtract 2 from 6 to get 4.
\frac{8\sqrt{2}\sqrt{6}-8\left(\sqrt{2}\right)^{2}}{4}
Use the distributive property to multiply 8\sqrt{2} by \sqrt{6}-\sqrt{2}.
\frac{8\sqrt{2}\sqrt{2}\sqrt{3}-8\left(\sqrt{2}\right)^{2}}{4}
Factor 6=2\times 3. Rewrite the square root of the product \sqrt{2\times 3} as the product of square roots \sqrt{2}\sqrt{3}.
\frac{8\times 2\sqrt{3}-8\left(\sqrt{2}\right)^{2}}{4}
Multiply \sqrt{2} and \sqrt{2} to get 2.
\frac{16\sqrt{3}-8\left(\sqrt{2}\right)^{2}}{4}
Multiply 8 and 2 to get 16.
\frac{16\sqrt{3}-8\times 2}{4}
The square of \sqrt{2} is 2.
\frac{16\sqrt{3}-16}{4}
Multiply -8 and 2 to get -16.