How do you factor #125x^9 + 729y^6#?
1 Answer
Jan 2, 2016
Use the sum of cubes identity to find:
#125x^9+729y^6 = (5x^3+9y^2)(25x^6-45x^3y^2+81y^4)#
Explanation:
Both
The sum of cubes identity can be written:
#a^3+b^3=(a+b)(a^2-ab+b^2)#
We can use this with
#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