Want to understand how a fraction is simplified

The fraction is used to determine the sum of a telescopic series.

$$\sum _{k=1}^{\mathrm{\infty}}\frac{1}{(k+1)\sqrt{k}+k\sqrt{k+1}}$$

This is the solved fraction.

$$\frac{1}{(k+1)\sqrt{k}+k\sqrt{k+1}}=\frac{(k+1)\sqrt{k}-k\sqrt{k+1}}{(k+1{)}^{2}k-{k}^{2}(k+1)}=\frac{(k+1)\sqrt{k}-k\sqrt{k+1}}{{k}^{3}+2{k}^{2}+k-{k}^{3}-{k}^{2}}=\frac{(k+1)\sqrt{k}-k\sqrt{k+1}}{k(k+1)}=\frac{\sqrt{k}}{k}-\frac{\sqrt{k+1}}{k+1}=\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}}$$

I want to understand what is being done on every step,mainly the last three. Thank you.

The fraction is used to determine the sum of a telescopic series.

$$\sum _{k=1}^{\mathrm{\infty}}\frac{1}{(k+1)\sqrt{k}+k\sqrt{k+1}}$$

This is the solved fraction.

$$\frac{1}{(k+1)\sqrt{k}+k\sqrt{k+1}}=\frac{(k+1)\sqrt{k}-k\sqrt{k+1}}{(k+1{)}^{2}k-{k}^{2}(k+1)}=\frac{(k+1)\sqrt{k}-k\sqrt{k+1}}{{k}^{3}+2{k}^{2}+k-{k}^{3}-{k}^{2}}=\frac{(k+1)\sqrt{k}-k\sqrt{k+1}}{k(k+1)}=\frac{\sqrt{k}}{k}-\frac{\sqrt{k+1}}{k+1}=\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}}$$

I want to understand what is being done on every step,mainly the last three. Thank you.