Factor.
-$64 s ^ { 3 } + 1$
A) $( 64 \mathrm {~s} + 1 ) \left( \mathrm { s } ^ { 2 } - 4 \mathrm {~s} + 1 \right)$
B) $( 4 \mathrm {~s} + 1 ) \left( 16 \mathrm {~s} ^ { 2 } + 1 \right)$
C) $( 4 \mathrm {~s} + 1 ) \left( 16 \mathrm {~s} ^ { 2 } - 4 \mathrm {~s} + 1 \right)$
D) $( 4 \mathrm {~s} - 1 ) \left( 16 \mathrm {~s} ^ { 2 } + 4 \mathrm {~s} + 1 \right)$

Question

Quizplus · Accepted Answer

We can use the identity $a^3+b^3=(a+b)(a^2-ab+b^2)$ to factor the expression. Taking $a=4s$ and $b=1$, we have:
\begin{align*}
64s^3+1 &= (4s)^3+1^3\
&= (4s+1)((4s)^2-(4s)(1)+1) \
&= (4s+1)(16s^2-4s+1) \
\end{align*}
Therefore, the answer is choice C.

Factor - 64s3+164 s ^ { 3 } + 164s3+1 A) (64 s+1)(s2−4 s+1)( 64 \mathrm {~s} + 1 ) \left( \mathrm { s } ^ { 2 } - 4 \mathrm {~s} + 1 \right)(64 s+1)(s2−4 s+1)

Factor
- $64 s ^ { 3 } + 1$ A) $( 64 \mathrm {~s} + 1 ) \left( \mathrm { s } ^ { 2 } - 4 \mathrm {~s} + 1 \right)$