Factor completely. If the polynomial cannot be factored, write "prime."
-$27 g^{3}+125 h^{3}$
A) $(3 g-5 h)\left(9 g^{2}+25 h^{2}\right)$
B) $(3 g+5 h)\left(9 g^{2}-15 g h+25 h^{2}\right)$
C) $\left(3 g+5 h^{2}\right)\left(9 g^{2}-15 g h+25 h^{2}\right)$
D) $(3 g-5 h)\left(9 g^{2}+15 g h+25 h^{2}\right)$

Question

Quizplus · Accepted Answer

This polynomial is a sum of cubes, factoring as $(3g)^3 + (5h)^3$, which can be factored using the sum of cubes formula $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$, resulting in choice B: $(3g + 5h)(9g^2 - 15gh + 25h^2)$.

Factor Completely 27g3+125h327 g^{3}+125 h^{3}27g3+125h3 A) (3g−5h)(9g2+25h2)(3 g-5 h)\left(9 g^{2}+25 h^{2}\right)(3g−5h)(9g2+25h2) B) (3g+5h)(9g2−15gh+25h2)(3 g+5 h)\left(9 g^{2}-15 g h+25 h^{2}\right)(3g+5h)(9g2−15gh+25h2)

Factor Completely $27 g^{3}+125 h^{3}$ A) $(3 g-5 h)\left(9 g^{2}+25 h^{2}\right)$ B) $(3 g+5 h)\left(9 g^{2}-15 g h+25 h^{2}\right)$