At first, we need to show that there is only one solution for $ a^3+12a^2+49a+69=0 $ Notice that $ f'(a)=3a^2+24a+49 $ is always positive since the discriminant of this polynomial is always negative. Similarly, there is a unique solution for $ b^3-9b^2+28b-31=0 $ Finally, because we have the following identity: \[ (-1-a)^3+12(-1-a)^2+49(-1-a)+69=-(a^3-9a^2+28a-31) \] Therefore, we conclude that $ a+b=1. \blacksquare $

An alternative natural approach, mentioned to me yesterday by @Strudan_Borisov and @Marinchoo is to set $t = a + b$ and substitute $b=t-a$ in the second equation, then combine equations and cancel $a$-s.

$f(x)=x^3 +12x^2 + 49x + 69$ $x\rightarrow x-\frac{b}{3a}\implies x\rightarrow x-4\implies f(x-4)=x^3+x+1=p(x)$ $\implies p(x)=0$ at $a+4$. $g(x)=x^3-9x^2+28x-31 = 0$ $x\rightarrow x-\frac{b}{3a}\implies x\rightarrow x+3\implies f(x+3)=x^3+x-1=-p(-x)$ $\implies p(-x)=0$ at $b-3\implies p(x)=0$ at $3-b$. Now $p(x)=x^3+x+1=0$ at $a+4, 3-b$. Now let's check nature of the roots $p'(x)=3x^2+1=0\implies x=\frac{i}{\sqrt3}\implies \text{no turning point}$ $\implies \text{only one real root}$ Since both $a,b\in\mathbb{R}$ and there's only one real root $\implies a+4=3-b$ Hence $\boxed{a+b=-1}$