So one day Bombelli was trying to solve $$x^3=15x+4$$ then he got $$x=\sqrt[3]{2+\sqrt{-121}}+\sqrt[3]{2-\sqrt{-121}}.$$
How did he know that $\sqrt[3]{2+\sqrt{-121}}=2+\sqrt{-1}$ and $\sqrt[3]{2-\sqrt{-121}}=2-\sqrt{-1}$?
So that $x=4$?
I have tried using $(a+ib)^3=2+1i$ then $a^3-3ab^2=2$ and $b^3-3a^2b=11$.
But it seems that you need to guess the solution.
On
When a polynomial equation with integer coefficients arises in your research, e.g., $$x^3-15x-4=0\ ,$$ then you "naturally" check whether it has integer solutions. In the case at hand the constant $4$ indicates that such solutions would be from the set $$\{-4,-2, -1,1,2,4\}\ .$$ It is easily seen that $\pm1$ and $\pm2$ cannot come up with the large $15$, so that only $-4$ and $4$ have to be tried.
I suppose that Bombelli, instead of trying to solve the equation $x^3=15x+4$, actually created it, knowing from the start that $4$ is a solution. And, if there is a number $a+b\sqrt{-1}$, with $a,b\in\Bbb Z$ such that$$\left(a+b\sqrt{-1}\right)^3=2+\sqrt{-121}=2+11\sqrt{-1}$$then $a^3-3ab^2=2$ and $3a^2b-b^3=11$. But note that$$a^3-3ab^2=2\iff a(a^2-3b^2)=2$$and, since we are searching for integers, there aren't as many choices as that…