Solution $(a=-2, b=-2)$ and $(a=1-n, b=1+n)$ and permutations thereof solve the equation $a^3+b^3=8-6ab$: Therefore min$(a+b)=-4$, Max$(a+b)=2$. Details of the calculations: Let $a+b=S, ab=P\Longrightarrow a^3+b^3=(a+b)^3-3ab(a+b)=S^3-3PS$. The equation at issue becomes $$S^3-3PS=8-6P\Longrightarrow S^3-8=3P(S-2)\Longrightarrow(S-2)(S^2+2S+4)=3P(S-2)$$ $$\Longrightarrow S-2=0,P=k\Longrightarrow a+b=2, ab=k\Longrightarrow (2-b)b=k\Longrightarrow a+b=2, ab=k\Longrightarrow b= 1\pm\sqrt{1-k}$$Hence $S=2$,$P=1-n^2$$\Longrightarrow a=1-n, b=1+n, ab=1-n^2$ (Examples: $a=1, b=1$ and $a=-3, b=5$). Furthermore, $S^2+2S+4-3P=0\Longrightarrow S=-1\pm\sqrt{3(P-1)}\Longrightarrow P=4, S=-4$ Whence $a=-2,b=-2$ is also a solution.

$0=a^3+b^3+(-2)^3-3(a)(b)(-2) =(a+b-2)(\frac{1}{2} ( (a+2)^2 + (b+2)^2 + (a-b)^2))$. $a+b=2$ or $a = b = -2 \implies a+b=-4$.

Let $a, b$ be two real numbers that satisfy $a^2b + b^3 = 8-6ab $. Prove that$$-6\leq a + b\leq 2$$