Consider a=0, then from both equations, y=-x. So in this case the tangent of the cubic is identical to the line; but it isn't an asymptote.
Now consider a≠0.
(x+y)3 expands to x3+3x2y+3xy2+y3=x3+y3+3xy(x+y).
But x3+y3=3axy, so (x+y)3=3axy+3xy(x+y)=3xy(a+x+y).
As x (and hence y) becomes large, 3axy becomes relatively small and negligible by comparison so y3≈-x3, implying that y≈-x and x+y→0, therefore (x+y)3→0. Since (x+y)3=3xy(a+x+y)→0, a+x+y→0, making a+x+y=0 an asymptote, since xy→-x2, which is never zero, but has a large magnitude as a negative number, when x is large.