Factor.
-$p ^ { 6 } - 64 y ^ { 6 }$
A) $\left( \mathrm { p } ^ { 2 } - 4 \mathrm { y } ^ { 2 } \right) \left( \mathrm { p } ^ { 4 } + 4 \mathrm { p } ^ { 2 } \mathrm { y } ^ { 2 } + 16 \mathrm { y } ^ { 4 } \right)$
B) $( p + 2 y ) ( p - 2 y ) \left( p ^ { 2 } + 2 p y + 4 y ^ { 2 } \right) \left( p ^ { 2 } - 2 p y + 4 y ^ { 2 } \right)$
C) $\left( p ^ { 3 } + 8 y ^ { 3 } \right) \left( p ^ { 3 } - 8 y ^ { 3 } \right)$
D) $( p + 2 y ) ( p - 2 y ) \left( p ^ { 4 } + 4 p ^ { 2 } y ^ { 2 } + 16 y ^ { 4 } \right)$

Question

Quizplus · Accepted Answer

We can recognize this as the difference of cubes with $p^2$ and $4y^2$ as the two cube bases. Applying the formula for difference of cubes, we get:

$(p^2)^3 - (4y^2)^3 = (p^2 - 4y^2) (p^4 + 4p^2y^2 + 16y^4)$

We can further factor the second factor using the sum and difference of cubes:

$p^4 + 4p^2y^2 + 16y^4 = (p^2)^2 + 2(p^2)(2y^2) + (2y^2)^2 + 2p^2y^2 - (2y^2)^2 = (p^2 + 2y^2)^2 - (2y^2)^2 = (p^2 + 2y^2 + 2y)(p^2 + 2y^2 - 2y)$

Thus, the final factorization is:

$p^6 - 64y^6 = (p^2 - 4y^2)(p^2 + 2y^2 + 2y)(p^2 + 2y^2 - 2y) = (p+2y)(p-2y)(p^2+2py+4y^2)(p^2-2py+4y^2)$

Option B is the only one that matches this factorization.

Factor - p6−64y6p ^ { 6 } - 64 y ^ { 6 }p6−64y6

Factor
- $p ^ { 6 } - 64 y ^ { 6 }$