How do you factor x^6-7x^3-8? Algebra Polynomials and Factoring Factoring Completely 1 Answer Antoine Apr 18, 2015 Like this, x^6-7x^3-8 -= (x^3)^2 -7(x^3) -8 -= (x^3-8)(x^3+1) -= (x^3-(2)^3)(x^3+(1)^3) Recall that, a^3+b^3=(a+b)(a^2-ab+b^2) and a^3-b^3=(a-b)(a^2+ab+b^2) => (x-2)(x^2+2x+4)(x+1)(x^2-x+1) Answer link Related questions What is Factoring Completely? How do you know when you have completely factored a polynomial? Which methods of factoring do you use to factor completely? How do you factor completely 2x^2-8? Which method do you use to factor 3x(x-1)+4(x-1) ? What are the factors of 12x^3+12x^2+3x? How do you find the two numbers by using the factoring method, if one number is seven more than... How do you factor 12c^2-75 completely? How do you factor x^6-26x^3-27? How do you factor 100x^2+180x+81? See all questions in Factoring Completely Impact of this question 24450 views around the world You can reuse this answer Creative Commons License