How do you factor #125x^9 + 729y^6#?

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Answer 1 · 2016-01-02T09:03:09

Use the sum of cubes identity to find:

#125x^9+729y^6 = (5x^3+9y^2)(25x^6-45x^3y^2+81y^4)#

Both #125x^9 = (5x^3)^3# and #729y^6 = (9y^2)^3# are perfect cubes.

The sum of cubes identity can be written:

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

We can use this with #a=5x^3# and #b=9y^2# to find:

#125x^9+729y^6#

#=(5x)^3+(9y^2)^3#

#=(5x^3+9y^2)((5x^3)^2-(5x^3)(9y^2)+(9y^2)^2)#

#=(5x^3+9y^2)(25x^6-45x^3y^2+81y^4)#

That's as far as we can go with Real coefficients.

If you allow Complex coefficients then it can be further factored as:

#=(5x^3+9y^2)(5x^3+9 omega y^2)(5x^3+9 omega^2 y^2)#

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