+128731 We can actually use a trick here to factor this as a sum of cubes.
x^4 + y^6 = (x^4/3)^3 + (y^2)^3....so we have
[x^(4/3) + y^(2)] [x^(8/3) - x^(4/3)y^(2) + y^(4)]
However, the first term could also be factored as a sum of cubes......and we could probably continue this process forever!!!!
+23247 If you can use only real numbers, it has no factors; there are factors over the complex numbers.
Do you need it factored over the complex numbers?
+128731 We can actually use a trick here to factor this as a sum of cubes.
x^4 + y^6 = (x^4/3)^3 + (y^2)^3....so we have
[x^(4/3) + y^(2)] [x^(8/3) - x^(4/3)y^(2) + y^(4)]
However, the first term could also be factored as a sum of cubes......and we could probably continue this process forever!!!!
