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\frac{48\left(\sqrt{3}+\sqrt{5}\right)}{\left(\sqrt{3}-\sqrt{5}\right)\left(\sqrt{3}+\sqrt{5}\right)}
Rationalize the denominator of \frac{48}{\sqrt{3}-\sqrt{5}} by multiplying numerator and denominator by \sqrt{3}+\sqrt{5}.
\frac{48\left(\sqrt{3}+\sqrt{5}\right)}{\left(\sqrt{3}\right)^{2}-\left(\sqrt{5}\right)^{2}}
Consider \left(\sqrt{3}-\sqrt{5}\right)\left(\sqrt{3}+\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}.
\frac{48\left(\sqrt{3}+\sqrt{5}\right)}{3-5}
Square \sqrt{3}. Square \sqrt{5}.
\frac{48\left(\sqrt{3}+\sqrt{5}\right)}{-2}
Subtract 5 from 3 to get -2.
-24\left(\sqrt{3}+\sqrt{5}\right)
Divide 48\left(\sqrt{3}+\sqrt{5}\right) by -2 to get -24\left(\sqrt{3}+\sqrt{5}\right).
-24\sqrt{3}-24\sqrt{5}
Use the distributive property to multiply -24 by \sqrt{3}+\sqrt{5}.