How do you factor #x^6-124x^3-125#?

1 Answer
Feb 17, 2017

#x^6-124x^3-125 = (x+1)(x^2-x+1)(x-5)(x^2+5x+25)#

Explanation:

The difference of cubes identity can be written:

#a^3-b^3 = (a-b)(a^2+ab+b^2)#

The sum of cubes identity can be written:

#a^3+b^3 = (a+b)(a^2-ab+b^2)#

Hence we find:

#x^6-124x^3-125 = (x^6-125x^3)+(x^3-125)#

#color(white)(x^6-124x^3-125) = x^3(x^3-125)+1(x^3-125)#

#color(white)(x^6-124x^3-125) = (x^3+1)(x^3-125)#

#color(white)(x^6-124x^3-125) = (x^3+1^3)(x^3-5^3)#

#color(white)(x^6-124x^3-125) = (x+1)(x^2-x+1)(x-5)(x^2+5x+25)#