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:

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

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

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

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