How do you factor $y^3 - 125$ ?

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Answer 1 · 2015-12-21T22:02:33

Use the difference of cubes identity to find:

$y^3-125 = (y-5)(y^2+5y+25)$

The difference of cubes identity can be written:

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

Use this with $a=y$ and $b=5$ ...

$y^3-125$

$=y^3-5^3$

$=(y-5)(y^2+(y)(5)+5^2)$

$=(y-5)(y^2+5y+25)$

The remaining quadratic factor has a negative discriminant, showing that it has no linear factors with Real coefficients.

If we are allowed Complex coefficients then we can factor a little further:

$=(y-5)(y-5omega)(y-5omega^2)$

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

How do you factor y^3 - 125y^3 - 125?