How do you factor #a^3-2ab^2-b^3#?

1 Answer
George C.
Mar 3, 2018

#a^3-2ab^2-b^3#

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

#= (a+b)(a-(1/2+sqrt(5)/2)b)(a-(1/2-sqrt(5)/2)b)#

Explanation:

Given:

#a^3-2ab^2-b^3#

Note that this is homogeneous of degree #3#.

So consider the related cubic in one variable:

#x^3-2x-1#

This has rational zero #x=-1# so factor #x+1# ...

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

The remaining quadratic factors by completing the square:

#x^2-x-1 = (x-1/2)^2-5/4#

#color(white)(x^2-x-1) = (x-1/2)^2-(sqrt(5)/2)^2#

#color(white)(x^2-x-1) = ((x-1/2)-sqrt(5)/2)((x-1/2)+sqrt(5)/2)#

#color(white)(x^2-x-1) = (x-1/2-sqrt(5)/2)(x-1/2+sqrt(5)/2)#

Then converting this back to homogeneous polynomials we have:

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

#color(white)(a^3-2ab^2-b^3) = (a+b)(a-(1/2+sqrt(5)/2)b)(a-(1/2-sqrt(5)/2)b)#

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