Factor.
-$216 y ^ { 3 } z - z$
A) $z ( 6 y - 1 ) \left( 36 y ^ { 2 } + 6 y + 1 \right)$
B) $z ( 216 y - 1 ) \left( y ^ { 2 } + 6 y + 1 \right)$
C) $z ( 6 y - 1 ) \left( 36 y ^ { 2 } + 1 \right)$
D) $z ( 6 y + 1 ) \left( 36 y ^ { 2 } - 6 y + 1 \right)$

Question

Quizplus · Accepted Answer

The correct factorization is $z ( 6 y - 1 ) \left( 36 y ^ { 2 } + 6 y + 1 \right)$. Factoring out $z$ from both terms gives $z(216y^3 - 1)$. Recognizing $216y^3$ as $(6y)^3$ and $1$ as $1^3$, we can apply the difference of cubes formula $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$ with $a = 6y$ and $b = 1$.

Factor - 216y3z−z216 y ^ { 3 } z - z216y3z−z

Factor
- $216 y ^ { 3 } z - z$