Solve.
-$$t ^ { 3 } + 64$$
A) $( t - 4 ) \left( t ^ { 2 } + 4 t + 16 \right)$
B) $( t - 64 ) \left( t ^ { 2 } - 1 \right)$
C) $( t + 4 ) \left( t ^ { 2 } + 16 \right)$
D) $( t + 4 ) \left( t ^ { 2 } - 4 t + 16 \right)$

Question

Quizplus · Accepted Answer

The expression $t^3 + 64$ can be factored as a sum of cubes: $t^3 + 4^3$. The sum of cubes formula is $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$. Applying this formula with $a = t$ and $b = 4$, we get $(t + 4)(t^2 - 4t + 16)$.

Solve - t3+64t ^ { 3 } + 64t3+64 A) (t−4)(t2+4t+16)( t - 4 ) \left( t ^ { 2 } + 4 t + 16 \right)(t−4)(t2+4t+16)

Solve
- $t ^ { 3 } + 64$ A) $( t - 4 ) \left( t ^ { 2 } + 4 t + 16 \right)$