factor expressions involving large powers

Question

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)