Factor completely. If the polynomial cannot be factored, write ʺprime.ʺ
-$$27 a ^ { 3 } - 64 b ^ { 3 }$$
A) $\left( 3 a + 4 b ^ { 2 } \right) \left( 9 a ^ { 2 } - 12 a b + 16 b ^ { 2 } \right)$
B) $( 3 a - 4 b ) \left( 9 a ^ { 2 } + 16 b ^ { 2 } \right)$
C) $( 3 a - 4 b ) \left( 9 a ^ { 2 } + 12 a b + 16 b ^ { 2 } \right)$
D) $( 27 a - 4 b ) \left( a ^ { 2 } + 12 a b + 16 b ^ { 2 } \right)$

Question

Quizplus · Accepted Answer

This is a difference of cubes, factored as $(a - b)(a^2 + ab + b^2)$ where $a = 3a$ and $b = 4b$, resulting in $(3a - 4b)(9a^2 + 12ab + 16b^2)$.

Factor Completely 27a3−64b327 a ^ { 3 } - 64 b ^ { 3 }27a3−64b3

Factor Completely $27 a ^ { 3 } - 64 b ^ { 3 }$