How do you factor: $y = x^{3} + \frac{1}{8}$ $y= x^3+1/8$ ?

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Answer 1 · 2016-10-03T22:41:31

Sum of cubes.
$(x + \frac{1}{2}) (x^{2} - \frac{x}{2} + \frac{1}{4})$ $(x+1/2)(x^2-x/2 +1/4)$

It helps to know the common squares and cubes!
This expression is the sum of two cubes.
It factors as

$x^{3} - y^{3} = (x - y) (x^{2} + x y + y^{2})$ $x^3 -y^3 = (x-y)(x^2+xy+y^2)$

Following the same pattern:

$x^{3} + \frac{1}{8} = (x + \frac{1}{2}) (x^{2} - \frac{x}{2} + \frac{1}{4})$ $x^3 +1/8= (x+1/2)(x^2-x/2 +1/4)$

How do you factor: y=x3+18y= x^3+1/8 ?