Factor completely. If the polynomial cannot be factored, write ʺprime.ʺ
-$$p ^ { 6 } + q ^ { 24 } r ^ { 18 }$$
A) $\left( p + q ^ { 4 } r ^ { 3 } \right) \left( p - q ^ { 4 } r ^ { 3 } \right) \left( p ^ { 4 } - p ^ { 2 } q ^ { 8 } r ^ { 6 } + q ^ { 16 } r ^ { 12 } \right)$
B) $\left( \mathrm { p } ^ { 3 } + \mathrm { q } ^ { 12 } \mathrm { r } ^ { 9 } \right) \left( \mathrm { p } ^ { 3 } - \mathrm { q } ^ { 12 } \mathrm { r } ^ { 9 } \right)$
C) $\left( p + q ^ { 4 } r ^ { 3 } \right) \left( p - q ^ { 4 } r ^ { 3 } \right) \left( p ^ { 2 } + p q ^ { 4 } r ^ { 3 } + q ^ { 8 } r ^ { 6 } \right) \left( p ^ { 2 } - q ^ { 4 } r ^ { 3 } + q ^ { 8 } r ^ { 6 } \right)$
D) $\left( p ^ { 2 } + q ^ { 8 } r ^ { 6 } \right) \left( p ^ { 4 } - p ^ { 2 } q ^ { 8 } r ^ { 6 } + q ^ { 16 } r ^ { 12 } \right)$

Question

Quizplus · Accepted Answer

The given polynomial is a sum of cubes, as $p^6$ is the cube of $p^2$ and $q^{24}r^{18}$ is the cube of $q^8r^6$. Thus, we can use the factorization for sum of cubes: $$a^3+b^3=(a+b)(a^2-ab+b^2)$$ Applying this factorization to our polynomial, we obtain: $$p^6+q^{24}r^{18}=(p^2)^3+(q^8r^6)^3=(p^2+q^8r^6)((p^2)^2-(p^2)(q^8r^6)+(q^8r^6)^2)$$ Simplifying the second factor, we have $(p^2)^2-(p^2)(q^8r^6)+(q^8r^6)^2 = p^4-p^2q^8r^6+q^{16}r^{12}$. Thus, the final factorization is: $$p^6+q^{24}r^{18}=\boxed{	extbf{(D)} (p^2+q^8r^6)(p^4-p^2q^8r^6+q^{16}r^{12})}$$

Factor Completely p6+q24r18p ^ { 6 } + q ^ { 24 } r ^ { 18 }p6+q24r18

Factor Completely $p ^ { 6 } + q ^ { 24 } r ^ { 18 }$