How do you factor completely #x^3+1/8#?

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Answer 1 · 2016-03-03T21:12:25

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

Both #x^3# and #1/8 = (1/2)^3# are perfect cubes. So we can use the "sum of cubes" identity:

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

with #a=x# and #b=1/2# as follows...

#x^3+1/8#

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

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

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

The remaining quadratic factor has no simpler factors with Real coefficients, but we can factor it if we allow Complex coefficients:

#=(x+1/2)(x+1/2omega)(x+1/2omega^2)#

where #omega = -1/2+sqrt(3)/2i# is the primitive Complex cube root of #1#.