$$1-y^3-x^2+y^3x^8+y^4x^2-y^4x^8$$
is divisible by all three brackets, that is seen from three couplings of terms: $(1-y^3)+(y^4x^2-x^2)+(y^3x^8-y^4x^8)$ is divisible by $y-1$, $(1-x^2)+(y^4x^2-y^4x^8)+(y^3x^8-y^3)$ by $x-1$ and $(1-y^4x^8)+(y^4x^2-y^3)+(y^3x^8-x^2)$ by $x^2y-1$. Thus the answer is 6.
It may be easily found by studying the Newton hexagon: its sides and diagonals have prescribed directions.