How do I factor #8x^3-1# completely?

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Answer 1 · 2015-07-26T18:11:13

Factor 8x^3 - 1

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

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

Answer 2 · 2015-07-26T18:14:28

You use the formula for the difference of perfect cubes.

All you really have to do in order to factor this expression completely is use the formula for the difference of two perfect cubes

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

You can rewrite the original expression as

#8x^3 - 1 = (2x)^3 - 1^3#

This will get you

#(2x)^3 - 1^3 = (2x - 1)[(2x)^2 + 2x * 1 + 1^2]#

#(2x)^3 - 1^3 = (2x - 1) * (4x^2 + 2x + 1)#

The quadratic #4x^2 + 2x + 1# cannot be factored further without using complex numbers, which I'm not sure you're supposed to use.