Factor Expressions Involving Large Powers, Radicals, and Trig Functions

You can use substitution to factor expressions involving large powers, radicals, and trig functions

Large powers :  If all the powers of your variables are multiplies of the same number, that's a good time to use substitution.
x¹² - 7x⁶ - 8
Since the degree of x in each term is a multiple of 6, you can rewrite the expression as a function of x6 :
(x6)² - 7x⁶ - 8
and solve using substitution. Make a new variable A to represent x6 , and you'll have
A² - 7A - 8,
Which you can factor like this:
(A - 8)(A + 1).
Then you can substitute back the x6 .
(x⁶ + 8)(x⁶ + 1).
You're not done! Don't forget that you might not be done factoring. The first factor happens to be a difference of cubes and the second factor is a sum of cubes:

x⁶ - 8 = (x²)3 - 23
= (x² - 2)(x4 + 2x² + 4)

x⁶ + 1 = (x²)3 + 13
= (x² + 1)(x⁴ -x² + 1)
So your final factorization is:
(x² - 2)(x⁴ + 2x² + 4) = (x² + 1)(x⁴ - x² + 1)