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Use mathematical induction to prove that 

$$1^3 + 2^3 + 3^3 + ... + n^3 = \dfrac{n^2 (n + 1)^2}{4}$$ 

for all positive integers n.

$$P(n) = 1^3 + 2^3 + 3^3 + ... + n^3 = \dfrac{n^2 (n + 1)^2}{4}$$
$$P(k) = \dfrac{k^2 (k + 1)^2}{4}$$
For $n = 1$
$$P(1) = \dfrac{1^2 (1 + 1)^2}{4}$$
$$ = \dfrac{1(2^2)}{4} = \dfrac{4}{4} = 1$$
$$P(n+1) = 1^3 + 2^3 + 3^3 + ... + n^3 + (n+1)^3= \dfrac{n^2 (n + 1)^2}{4} + (n+1)^3$$
For $(k+1)$
$$P(k+1) = a_k + (k + 1)^3$$
$$ = \dfrac{k^2 (k + 1)^2}{4} + (k + 1)^3$$
$$ = \dfrac{k^2 (k + 1)^2+4(k + 1)^3}{4}$$
$$ = \dfrac{(k+1)^2[k^2+4(k + 1)]}{4}$$
$$ = \dfrac{(k+1)^2(k^2+4k + 4)}{4}$$
$$ = \dfrac{(k+1)^2(k+2)^2}{4}$$
Let $w = k+1$
$$ = \dfrac{w^2(w+1)^2}{4} \square$$