How do you factor $p^{6} - 1$ $p^6 - 1$ ?

Question

3614 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-12-08T03:23:33

$(p + 1) (p - 1) (p^{2} - p + 1) (p^{2} + p + 1)$ $(p+1)(p-1)(p^2-p+1)(p^2+p+1)$

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

How do you factor p^6 - 1p6−1p^6 - 1?