Factor completely. If the polynomial cannot be factored, write ʺprime.ʺ
-$$64 g ^ { 3 } + 125 h ^ { 3 }$$
A) $( 4 \mathrm {~g} + 5 \mathrm {~h} ) \left( 16 \mathrm {~g} ^ { 2 } - 20 \mathrm { gh } + 25 \mathrm {~h} ^ { 2 } \right)$
B) $( 4 g - 5 h ) \left( 16 g ^ { 2 } + 25 h ^ { 2 } \right)$
C) $\left( 4 g + 5 h ^ { 2 } \right) \left( 16 g ^ { 2 } - 20 g h + 25 h ^ { 2 } \right)$
D) $( 4 g - 5 h ) \left( 16 g ^ { 2 } + 20 g h + 25 h ^ { 2 } \right)$

Question

Quizplus · Accepted Answer

This is a sum of two cubes, which can be factored using the formula: $a^3+b^3=(a+b)(a^2-ab+b^2)$. Applying this formula with $a=4g$ and $b=5h$, we get $(4g+5h)(16g^2-20gh+25h^2)$.

Factor Completely 64g3+125h364 g ^ { 3 } + 125 h ^ { 3 }64g3+125h3

Factor Completely $64 g ^ { 3 } + 125 h ^ { 3 }$