How do you factor #1-124x^3#?

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Answer 1 · 2015-06-27T22:11:22

Use the difference of cubes identity:

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

Let #alpha = root(3)(124)#

Then #1-124x^3 = (1-alpha x)(1+ alpha x + alpha^2 x^2)#

I think the #124# in the question should have been #125#, but I will answer the question as it stands.

Let #alpha = root(3)(124)#

Then:

#1 - 124x^3#

#= 1^3 - (alpha x)^3#

#= (1 - alpha x)(1 + alpha x + (alpha x)^2)#

#= (1 - alpha x)(1 + alpha x + alpha^2 x^2)#

The remaining quadratic factor has no simpler linear factors with real coefficients.

If the #124# was #125 = 5^3# then we would have

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