Factor completely.
-$$n ^ { 3 } - \frac { 1 } { 64 }$$
A) $\left( n - \frac { 1 } { 4 } \right) \left( n ^ { 2 } + \frac { 1 } { 4 } n + \frac { 1 } { 16 } \right)$
B) $\left( n + \frac { 1 } { 16 } \right) \left( n ^ { 2 } - \frac { 1 } { 4 } n + \frac { 1 } { 4 } \right)$
C) $\left( n - \frac { 1 } { 16 } \right) \left( n ^ { 2 } + \frac { 1 } { 4 } n + \frac { 1 } { 4 } \right)$
D) $\left( n + \frac { 1 } { 4 } \right) \left( n ^ { 2 } - \frac { 1 } { 4 } n + \frac { 1 } { 16 } \right)$

Question

Quizplus · Accepted Answer

The expression $n^3 - \frac{1}{64}$ is a difference of cubes, which can be factored as $(a^3 - b^3) = (a - b)(a^2 + ab + b^2)$. Here, $a = n$ and $b = \frac{1}{4}$, so the factored form is $\left(n - \frac{1}{4}\right)\left(n^2 + \frac{1}{4}n + \frac{1}{16}\right)$.

Factor Completely n3−164n ^ { 3 } - \frac { 1 } { 64 }n3−641​

Factor Completely $n ^ { 3 } - \frac { 1 } { 64 }$