How do you factor #p^6 - 1#?

1 Answer
Dec 8, 2015

Answer:

#(p+1)(p-1)(p^2-p+1)(p^2+p+1)#

Explanation:

Identities to recognize: #{((a^2-b^2)=(a+b)(a-b)),((a^3+b^3)=(a+b)(a^2-ab+b^2)),((a^3-b^3)=(a-b)(a^2+ab+b^2)):}#

Through the first identity, a difference of squares, #p^6-1=(p^3+1)(p^3-1)#.

Now, we have the other two identities—a sum of cubes and a difference of cubes.

#color(blue)((p^3+1))color(magenta)((p^3-1))=color(blue)((p+1)(p^2-p+1))color(magenta)((p-1)(p^2+p+1))#

Thus, #p^6-1=(p+1)(p-1)(p^2-p+1)(p^2+p+1)#.