Sum of the cubes of first n natural numbers = $\left[\dfrac{n × (n + 1)}{2}\right]^{2}$
Sum of the cubes of first 12 numbers = $\left[\dfrac{12 × (12 + 1)}{2}\right]^{2}$
= $\left[\dfrac{12 × 13}{2}\right]^{2}$
= $(6 × 13)^{2}$
= $78^{2}$
= 6084