Rationalise the denominator of $\frac{5}{\sqrt{3}-\sqrt{5}}$.

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Here we use the Identity $(iii)$ given earlier.

So, $\frac{5}{\sqrt{3}-\sqrt{5}}=\frac{5}{\sqrt{3}-\sqrt{5}} \times \frac{\sqrt{3}+\sqrt{5}}{\sqrt{3}+\sqrt{5}}=\frac{5(\sqrt{3}+\sqrt{5})}{3-5}=\left(\frac{-5}{2}\right)(\sqrt{3}+\sqrt{5})$

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$(ii)$ $(3+\sqrt{3})(3-\sqrt{3})$

$(iii)$ $(\sqrt{5}+\sqrt{2})^{2}$

$(iv)$ $(\sqrt{5}-\sqrt{2})(\sqrt{5}+\sqrt{2})$

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$(iii)$ $16^{\frac{3}{4}}$

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