Prove the following by using the principle of mathematical induction for all $n \in N$ :
$1^{3}+2^{3}+3^{3}+\ldots+n^{3}=\left(\frac{n(n+1)}{2}\right)^{2}$

Let the given statement be $P(n),$ i.e.,
$P(n): 1^{3}+2^{3}+3^{3}+\ldots+n^{3}=\left(\frac{n(n+1)}{2}\right)^{2}$
For $n=1,$ we have
$P(1): 1^{3}=1=\left(\frac{1(1+1)}{2}\right)^{2}=\left(\frac{2}{2}\right)^{2}=1^{2}=1,$ which is true.
Let $P(k)$ be true for some positive integer $k,$ i.e.,
$1^{3}+2^{3}+3^{3}+\ldots+k^{3}=\left(\frac{k(k+1)}{2}\right)^{2}$ ........$(i)$
We shall now prove that $P(k+1)$ is true.
Consider
$1^{3}+2^{3}+3^{3}+\ldots+k^{3}+(k+1)^{3}$
$= \left(1^{3}+2^{3}+3^{3}+\ldots+k^{3}\right)+(k+1)^{3}$
$= \left(\frac{k(k+1)}{2}\right)^{2}+(k+1)^{3}$ [Using $(i)$]
$= \frac{k^{2}(k+1)^{2}}{4}+(k+1)^{3}$
$= \frac{k^{2}(k+1)^{2}+4(k+1)^{3}}{4}$
$= \frac{(k+1)^{2}\{k^{2}+4(k+1)\}}{4}$
$= \frac{(k+1)^{2}\{k^{2}+4k+4\}}{4}$
$= \frac{(k+1)^{2}(k+2)^{2}}{4}$
$= \left(\frac{(k+1)(k+2)}{2}\right)^{2}$
Thus,$P(k+1)$ is true whenever $P(k)$ is true.
Hence,by the principle of mathematical induction,the statement $P(n)$ is true for all natural numbers $n \in N.$