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

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Answer 1 · 2015-04-09T20:45:12

This problem involves factoring the sum of cubes.

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

For the expression #(x^3+1/8)#:

#a=root3(x^3)=x#

#b=root3(1/8)=1/2#

Solution:

#(x^3+1/8)# =

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

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