How do you factor #(x^6)+(7x^3)-8#?

1 Answer
Feb 3, 2017

Answer:

#x^6+7x^3-8 = (x+2)(x^2-2x+4)(x-1)(x^2+x+1)#

Explanation:

The sum of cubes identity can be written:

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

The difference of cubes identity can be written:

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

So we find:

#x^6+7x^3-8 = (x^3+8)(x^3-1)#

#color(white)(x^6+7x^3-8) = (x^3+2^3)(x^3-1^3)#

#color(white)(x^6+7x^3-8) = (x+2)(x^2-2x+4)(x-1)(x^2+x+1)#