Find the Product (n-4)^3

${(n - 4)}^{3}$

Use the Binomial Theorem.

$n^{3} + 3 n^{2} \cdot - 4 + 3 n {(- 4)}^{2} + {(- 4)}^{3}$

Simplify each term.

Multiply $- 4$ by $3$ .

$n^{3} - 12 n^{2} + 3 n {(- 4)}^{2} + {(- 4)}^{3}$

Raise $- 4$ to the power of $2$ .

$n^{3} - 12 n^{2} + 3 n \cdot 16 + {(- 4)}^{3}$

Multiply $16$ by $3$ .

$n^{3} - 12 n^{2} + 48 n + {(- 4)}^{3}$

Raise $- 4$ to the power of $3$ .

$n^{3} - 12 n^{2} + 48 n - 64$