Factor completely. $64 m ^ { 6 } - n ^ { 6 } \quad$ [Hint: $64 = 2 ^ { 6 }$.]
A) $( 2 m - n ) \left( 4 m ^ { 2 } + 2 m n + n ^ { 2 } \right) \left( 8 m ^ { 3 } + n ^ { 3 } \right)$
B) $( 2 m + n ) \left( 4 m ^ { 2 } - 2 m n + n ^ { 2 } \right) ( 2 m - n ) \left( 4 m ^ { 2 } + 2 m n + n ^ { 2 } \right)$
C) $\left( 8 m ^ { 3 } - n ^ { 3 } \right) \left( 8 m ^ { 3 } + n ^ { 3 } \right)$
D) Prime

Question

Quizplus · Accepted Answer

The expression $64 m^6 - n^6$ can be factored as a difference of squares: $(8m^3)^2 - (n^3)^2$, which factors to $(8m^3 - n^3)(8m^3 + n^3)$. Each of these can be further factored using the sum and difference of cubes formulas.

Factor Completely 64m6−n664 m ^ { 6 } - n ^ { 6 } \quad64m6−n6

Factor Completely $64 m ^ { 6 } - n ^ { 6 } \quad$