Roman C. answered 05/08/13
Every perfect square trinomial has the form
x2n + 2axn + a2.
where n is a positive integer.
It can factor as (xn + a)2
Difference of two nth powers:
n = 2 → a2 - b2 = (a - b)(a + b) Squares.
n = 3 → a3 - b3 = (a - b)(a2 + ab + b2) Cubes.
n = 4 → a4 - b4 = (a - b)(a3 + a2b + ab2 + b3).
n = 5 → a5 - b5 = (a - b)(a4 + a3b + a2b2 + ab3 + b4).
Can you see the pattern?
For odd n you can also factor sums of two nth powers
n = 3 → a3 + b3 = (a + b)(a2 - ab + b2) Cubes.
n = 5 → a5 + b5 = (a + b)(a4 - a3b + a2b2 - ab3 + b4).
n = 7 → a7 + b7 = (a + b)(a6 - a5b + a4b2 - a3b3 + a2b4 - ab5 +b6).
Can you see the pattern?
Also, for both sums and differences of two nth powers:
1. If n is prime, the second factor is a prime polynomial and so no further factorization can be done.
2. If n is composite, the second factor can be factored using factoring by grouping.