A cubic equation $$x^3+ax^2+bx+c=0$$ has three solutions, which can be found analytically.
Likewise, a vector equation like $$\underline{\underline{A}} \underline{x} + c(\underline{x}^T\underline{x})\underline{x} + \underline{b} = 0$$ should have three vector solutions $\underline{x}_0,\underline{x}_1,\underline{x}_2$. My question is, how do I find these? I would appreciate analytical ideas, but numerical solutions would also be fine, the only problem with the latter is that I need all three solution (or how many real solutions there are) rendering (at least in my mind) iterative methods infeasible.
If it helps we can assume $\underline{\underline{A}} = \text{diag} \underline{a}$.
Would appreciate any help! Thanks.
Hint: Expanding the vector equation and using the assumption for $A$, note that you get 3 cubic equations:
$$a_0 \, x_0+ c \, (x_0^2+x_1^2+x_2^2)\, x_0 + b_0 = 0$$ $$a_1 \, x_1+ c \, (x_0^2+x_1^2+x_2^2)\, x_1 + b_1 = 0$$ $$a_2 \, x_2+ c \, (x_0^2+x_1^2+x_2^2)\, x_2 + b_2 = 0$$
you can solve this system of equations traditionally.