Polynomial of cubes.

Let us define another function:

$g(x) = f(x) - x^{3}$

Thus $g(x) = 0$ , for 'x' being any integer from 1 to 10

hence $(x-1), (x-2), ... (x-10)$ divides $g(x)$

thus, $g(x) = c*(x-1)*(x-2)*..(x-10)$

As, $f(11) = 1332$

$g(11) = f(11) - 11^{3}$

$g(11) = 1332 - 1331 = 1$

$c*(11-10)*(11-9)*...(11-1) = 1$

$c = \frac{1}{10!}$

$g(x) = \frac{(x-1)*(x-2)*...(x-10)}{10!}$

Thus, $f(x) = \frac{(x-1)*(x-2)*...(x-10)}{10!} + x^{3}$

Hence,

$f(12) = \frac{11*10*9*...2}{10!} + 12^{3} = \frac{11!}{10!} + 1728 = \boxed{1739}$