How do you factor #x^3y^6 + 8#?
1 Answer
Dec 14, 2015
Explanation:
The sum of cubes identity can be written:
#a^3+b^3=(a+b)(a^2-ab+b^2)#
Both
#x^3y^6+8#
#=(xy^2)^2+2^3#
#=(xy^2+2)((xy^2)^2-(xy^2)(2)+2^2)#
#=(xy^2+2)(x^2y^4-2xy^2+4)#
That's as far as you can get with Real coefficients, but if you allow Complex coefficients this factors a little further:
#=(xy^2+2)(xy^2+2omega)(xy^2+2omega^2)#
where