Factor the polynomial completely
-$27 y^{12}+8$
A) $27 y^{12}+9$
B) $\left(3 y^{4}+2\right)^{3}$
C) $\left(9 y^{4}-4\right)\left(3 y^{4}+2\right)$
D) $\left(3 y^{4}+2\right)\left(9 y^{8}-6 y^{4}+4\right)$

Question

Quizplus · Accepted Answer

The given polynomial is a sum of two cubes: $27y^{12}+8 = (3y^4)^3 + 2^3$. Factoring a sum of cubes uses the formula $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$, resulting in $(3y^4 + 2)((3y^4)^2 - 3y^4*2 + 2^2)$, which simplifies to $(3y^4 + 2)(9y^8 - 6y^4 + 4)$.

Factor the Polynomial Completely - 27y12+827 y^{12}+827y12+8 A) 27y12+927 y^{12}+927y12+9 B) (3y4+2)3\left(3 y^{4}+2\right)^{3}(3y4+2)3

Factor the Polynomial Completely
- $27 y^{12}+8$ A) $27 y^{12}+9$ B) $\left(3 y^{4}+2\right)^{3}$