Let the given statement be P(n) i.e., P (n) : $1^3+2^3+3^3$ + ... + $n^3$ = $\left(\frac{n(n+1)}{2}\right)^2$ First we prove that the statement is true for n = 1. P (1) : = $\left(\frac{1(1+1)}{2}\right)^2$ = $\left(\frac{1\times 2}{2}\right)^2$ = $(1)^2$ = 1 , which is true Assume that P(k) is true for some positive integer k, i.e., $1^3+2^3+3^3$ + ... + $k^3$ = $\left(\frac{k(k+1)}{2}\right)^2$ ... (i) We shall now prove that P(k + 1) is also true. For this we have to prove $1^3+2^3$ + ... + $k^3+(k+1)^3$ = $\left(\frac{(k+1)[(k+1)+1]}{2^2}\right)$ L.H.S = $1^3+2^3$ + ... + $k^3+(k+1)^3$ = $\left(\frac{k(k+1)}{2}\right)^2$ + $(k+1)^3$ (By using (i)) = $(k+1)^2\left(\frac{k^2}{4}+k+1\right)$ = $(k+1)^2\left(\frac{k^2+4k+4}{4}\right)$ = $(k+1)^2\left(\frac{(k+2)^2}{2}\right)$ = $\frac{(k+1)^2((k+1)+1)^2}{2^2}$ = $\left(\frac{(k+1)[(k+1)+1]}{2^2}\right)$ ∴ P(k + 1) is true, whenever P(k) is true. Hence, by the principle of mathematical induction, P(n) is true ∀ n ∈ N.