General solution to the equation $$(x^3 + 3xy^2)dx + (3x^2y + y^3)dy = 0$$ is (c is a constant)

A

$$\frac {1}{4}(x^4 + 6x^2y^2 + y^4) = c$$

B

$$\frac {1}{5}(8x^4 + 6x^3 y + y^3) = c$$

C

$$\frac {1}{12}(x^3 + 6xy^2 + y^4) = c$$

D

$$(x^2 + y^2) = c$$