How do you factor x^3-1?

x^3 - 1

This is the difference of two cubes. Use the formulas found here:
http://www.purplemath.com/modules/specfact2.htm

a^3 – b^3 = (a – b)(a^2 + ab + b^2)

a = x

b = 1

x^3 - 1 = (x - 1)(x^2 + x + 1)

x^3 - 1 = (x-1)(x^2+x+1)

Heres a good little trick thats proven to work.

Think of a number you can put into that equation for x and get it to equal 0. 1 is the first number you can think of for this since 1^3-1 = 1-1 = 0.

Then a factor of that polynomial is (x - 1)

This is called the factor theorem, generally for any polynomial f(x) if f(a) = 0 then (x-a) is a factor.

Your final answer is (x-1)(x^2 + x +1) and the latter doesnt factor down any more as it has no real roots.

Hope this helps.

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

x^3 - 1
= x^3 - 1^3
= (x - 1)(x^2 + x + 1^2)
= (x - 1)(x^2 + x + 1)

This is called a difference of a cube because both numbers are perfect cubes, and you're subtracting. There is a formula you can memorize so you don't have to try and factor it: (A-B)(A^2+AB+B^2). For your problem, A is x and B is 1. (x-1)(x^2+x+1).

(x-1)(x^2+x+1)

use factors of sum of cubes:

a^3 + b^3 = (a + b)(a^2 – ab + b^2)

to get

(x - 1)(x^2 + x + 1)












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