Factor.
-$343 y^{3} z-z$
A) $z(7 y+1)\left(49 y^{2}-7 y+1\right)$
B) $z(7 y-1)\left(49 y^{2}+7 y+1\right)$
C) $z(343 y-1)\left(y^{2}+7 y+1\right)$
D) $z(7 y-1)\left(49 y^{2}+1\right)$

Question

Quizplus · Accepted Answer

The expression can be factored by first factoring out the common factor $z$, resulting in $z(343y^3 - 1)$. Recognizing $343y^3$ as $(7y)^3$ and $1$ as $1^3$, we can apply the difference of cubes formula $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$ where $a = 7y$ and $b = 1$. This gives us $z(7y - 1)(49y^2 + 7y + 1)$, which matches option B.

Factor - 343y3z−z343 y^{3} z-z343y3z−z A) z(7y+1)(49y2−7y+1)z(7 y+1)\left(49 y^{2}-7 y+1\right)z(7y+1)(49y2−7y+1) B) z(7y−1)(49y2+7y+1)z(7 y-1)\left(49 y^{2}+7 y+1\right)z(7y−1)(49y2+7y+1)

Factor
- $343 y^{3} z-z$ A) $z(7 y+1)\left(49 y^{2}-7 y+1\right)$ B) $z(7 y-1)\left(49 y^{2}+7 y+1\right)$