How do you factor $\frac{1}{8} x^{3} - \frac{1}{27} y^{3}$ $1/8x^3-1/27y^3$ ?

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Answer 1 · 2015-05-21T21:52:05

$\frac{1}{8} x^{3} - \frac{1}{27} y^{3}$ $1/8x^3 - 1/27y^3$

$= {(\frac{1}{2})}^{3} x^{3} - {(\frac{1}{3})}^{3} y^{3}$ $= (1/2)^3x^3-(1/3)^3y^3$

$= {(\frac{1}{2} x)}^{3} - {(\frac{1}{3} y)}^{3}$ $= (1/2x)^3-(1/3y)^3$

$= {(\frac{x}{2})}^{3} - {(\frac{y}{3})}^{3}$ $= (x/2)^3-(y/3)^3$

This is of the form $(a^{3} - b^{3})$ $(a^3-b^3)$ which has a well known factorization:

$a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ $a^3-b^3 = (a-b)(a^2+ab+b^2)$

So we can substitute $a = \frac{x}{2}$ $a = x/2$ , $b = \frac{y}{3}$ $b = y/3$ to get

${(\frac{x}{2})}^{3} - {(\frac{y}{3})}^{3}$ $(x/2)^3-(y/3)^3$

$= (\frac{x}{2} - \frac{y}{3}) ({(\frac{x}{2})}^{2} + (\frac{x}{2}) (\frac{y}{3}) + {(\frac{y}{3})}^{2})$ $= (x/2-y/3)((x/2)^2+(x/2)(y/3)+(y/3)^2)$

$= (\frac{x}{2} - \frac{y}{3}) (\frac{x^{2}}{4} + \frac{x y}{6} + \frac{y^{2}}{9})$ $= (x/2-y/3)(x^2/4+(xy)/6+y^2/9)$

This is as far as we can go with real valued coefficients.

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