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#.