Question

Find factors of p^6+p^3-p^2+p-1. Explain in detail.

Answer 1

p^6 + p³ - p² +p -1

p³(p³+1) - p² + p - 1

p³(p³+1³) - p² + p - 1

p³ (p+1) (p²-p+1) - 1 (p²-p+1)

(p²-p+1) [p³(p+1) -1]

(p²-p+1) (p^4+p³-1)

this is your answer

I used identity a³+b³ = (a+b) ( a²+b²-ab) in (p³+1)

1 can also be written as 1³

$(p^{2} - p + 1) \: (p^{4} + p^{3} - 1)$

factors