Ta có
$1^3 = 1^2$
$1^3 + 2^3 = 9 = 3^2 = (1+2)^2 = (\dfrac{2.3}{2})^2$
$1^3 + 2^3 + 3^3 = 36 = 6^2 = (1+2+3)^2 = (\dfrac{3.4}{2})^2$
$1^3 + 2^3 + 3^3 + 4^3 = 100 = 10^2 = (1+2+3+4)^2 = (\dfrac{4.5}{2})^2$
...
$1^3 + 2^3 + \cdots + n^3 = (1+2+\cdots + n)^2 = (\dfrac{n(n+1)}{2})^2$
Vậy ta có
$1^3 + 2^3 + \cdots + n^3 = (\dfrac{n(n+1)}{2})^2$