| Solution by the proposer.
If the cubic is to factor in C[x,y], then one of the factors must be
linear. Without loss of generality, we may assume this factor is of
the form x-py-q, where p and q are complex numbers.
Substituting x=py+q in the original cubic, we get a polynomial in y
that must be identically 0. Thus each of its coefficients must be 0.
This gives us the four equations:
1 + p + p^3 = 0
4 + 2p + 3p^2 + q + 3 p^2 q = 0
3 + aq + 3q^2 + q^3 = 0
b + ap + 2q + 6pq + 3 p q^2 = 0
Solving these equations simultaneously, yields a=4 and b=5.
As a check we note that the resulting polynomial can be written as
(x+y+2)(y+1)^2 + (x+1)^3. The reducibility of this polynomial will not
change if we let x=X-1 and y=Y-1. This produces the polynomial X^3 +
XY^2 + Y^3. Letting z=X/Y shows that this polynomial factors over
C[x,y] since z^3+z+1 factors of C[z].
Rabinowitz also remarks that,
"except for a few exceptional cases, if f(x,y) is a cubic polynomial in
C[x,y], there will be unique complex constants a and b such that
f(x,y)+ax+by factors over C[x,y]."
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