Factor.
-$p ^ { 6 } + 1$
A) $\left( \mathrm { p } ^ { 3 } + 1 \right) \left( \mathrm { p } ^ { 3 } - 1 \right)$
B) $( p + 1 ) ( p - 1 ) \left( p ^ { 2 } + p + 1 \right) \left( p ^ { 2 } - p + 1 \right)$
C) $\left( p ^ { 2 } + 1 \right) \left( p ^ { 4 } - p ^ { 2 } + 1 \right)$
D) $( p + 1 ) ( p - 1 ) \left( p ^ { 4 } - p ^ { 2 } + 1 \right)$

Question

Quizplus · Accepted Answer

We can use the identity for the sum of two cubes:

$$a^3+b^3=(a+b)(a^2-ab+b^2)$$

Letting $p^2=x,$ we can rewrite the expression as:

$$p^6+1=x^3+1=(x+1)(x^2-x+1)$$

Now, we can substitute back $x=p^2$ to get:

$$p^6+1=(p^2+1)\left(p^4-p^2+1ight)$$

Therefore, the correct choice is $\boxed{	extbf{(C)}}.$

Factor - p6+1p ^ { 6 } + 1p6+1 A) (p3+1)(p3−1)\left( \mathrm { p } ^ { 3 } + 1 \right) \left( \mathrm { p } ^ { 3 } - 1 \right)(p3+1)(p3−1)

Factor
- $p ^ { 6 } + 1$ A) $\left( \mathrm { p } ^ { 3 } + 1 \right) \left( \mathrm { p } ^ { 3 } - 1 \right)$