Find the value

$=x^{3}-12 x^{2}+48 x-64$

$=x^{3}-3 \times x^{2} \times 4+3 \times 4^{2} \times x-4^{3}$

$=(x-4)^{3}$

$\left[\therefore a^{3}-b^{3}-3 a^{2} b+3 a b^{2}=(a-b)^{3}\right]$

$=(x-4)(x-4)(x-4)$

$\therefore x^{3}-12 x(x-4)-64=(x-4)(x-4)(x-4)$