How do you factor completely #x^3 + y^3#?

1 Answer
Mar 30, 2018

Answer:

#x^3+y^3 = (x+y)(x^2-xy+y^2)#

#color(white)(x^3+y^3) = (x+y)(x-(1/2+sqrt(3)/2i)y)(x-(1/2-sqrt(3)/2i)y)#

Explanation:

Given:

#x^3+y^3#

Note that if #y = -x# then this is zero. So we can deduce that #(x+y)# is a factor and separate it out:

#x^3+y^3=(x+y)(x^2-xy+y^2)#

We can calculate the discriminant for the remaining homogeneous quadratic in #x# and #y# just like we would for a quadratic in a single variable:

#x^2-xy+y^2#

is in standard form:

#ax^2+bxy+cy^2#

with #a=1#, #b=-1# and #c=1#

This has discriminant #Delta# given by the formula:

#Delta = b^2-4ac = color(blue)(1)^2-4(color(blue)(1))(color(blue)(1)) = -3#

Since #Delta < 0#, this quadratic has no linear factors with real coefficients.

We can factor it with complex coefficients by completing the square and using #i^2=-1# as follows:

#x^2-xy+y^2 = (x-1/2y)^2+3/4y^2#

#color(white)(x^2-xy+y^2) = (x-1/2y)^2+(sqrt(3)/2 y)^2#

#color(white)(x^2-xy+y^2) = (x-1/2y)^2-(sqrt(3)/2 y i)^2#

#color(white)(x^2-xy+y^2) = ((x-1/2y)-sqrt(3)/2i y)((x-1/2y)+sqrt(3)/2i y)#

#color(white)(x^2-xy+y^2) = (x-(1/2+sqrt(3)/2i)y)(x-(1/2-sqrt(3)/2i)y)#

So:

#x^3+y^3 = (x+y)(x^2-xy+y^2)#

#color(white)(x^3+y^3) = (x+y)(x-(1/2+sqrt(3)/2i)y)(x-(1/2-sqrt(3)/2i)y)#