Factor completely. If the polynomial cannot be factored, write ʺprime.ʺ
-$125 \mathrm { x } ^ { 12 } + 8 \mathrm { y } ^ { 9 }$
A) $\left( 5 x ^ { 4 } + 2 y ^ { 3 } \right) \left( 25 x ^ { 8 } + 10 x ^ { 4 } y ^ { 3 } + 4 y ^ { 6 } \right)$
B) $\left( 5 x ^ { 4 } - 2 y ^ { 3 } \right) \left( 5 x ^ { 4 } + 2 y ^ { 3 } \right)$
C) $\left( 5 x ^ { 4 } - 2 y ^ { 3 } \right) \left( 25 x ^ { 2 } + 10 x ^ { 4 } y ^ { 3 } + 4 y ^ { 6 } b ^ { 2 } \right)$
D) $\left( 5 x ^ { 4 } + 2 y ^ { 3 } \right) \left( 25 x ^ { 8 } - 10 x ^ { 4 } y ^ { 3 } + 4 y ^ { 6 } \right)$

Question

Quizplus · Accepted Answer

We can use the formula for the sum of cubes to factor $125 x^{12}+8y^9.$ This formula states that $a^3+b^3 = (a+b)(a^2-ab+b^2).$
We have $125x^{12}=(5x^4)^3$ and $8y^9=(2y^3)^3,$ so we can apply the formula as follows: $$125x^{12}+8y^9 = (5x^4)^3+(2y^3)^3 = (5x^4+2y^3)(25x^8-10x^4y^3+4y^6).$$ Therefore, the answer is D.

Factor Completely 125x12+8y9125 \mathrm { x } ^ { 12 } + 8 \mathrm { y } ^ { 9 }125x12+8y9

Factor Completely $125 \mathrm { x } ^ { 12 } + 8 \mathrm { y } ^ { 9 }$