Are you smarter than me, Carl?

Since we only require two decimal places (which is very little), Intermediate Value Theorem is preferred. While Cardano's Method looks messy and Newton Raphson Method requires a background in calculus and plenty of tedious work.

Let $f(x) = x^3 + 12x - 12$ . Because $f(0) < 0, f(1) > 0$ , the root is in between $0$ and $1$ exclusive.

$f(0.5) < 0, f(0.7) < 0, f(0.9) < 0, f(0.95) > 0, \\ f(0.93) < 0, f(0.94) > 0, f(0.935) > 0$

Since $f(0.93) < 0 < f(0.935)$ , the root of the equation $x^3 + 12x - 12 = 0$ is equals to $\boxed{0.93}$ to 2 decimal places.

And suppose there is another root, by fermat's theorem, there exist a point $c$ such that $f'(c) = 0$ but since $f'(x) = 3x^2 + 12 > 0$ for all $x$ , there can't be any more roots.

Match the equation with the identity $(u+v)^3-3uv(u+v)-(u^3+v^3)$ .We get the following system of equations: $\begin{cases}u+v=x\\-3uv=12\rightarrow uv=-4\\u^3+v^3=12\end{cases}$ Cube the second equation to get $u^3v^3=-64$ .Make an equation in a variable $z$ having roots $u^3,v^3$ : $(z-u^3)(z-v^3)=z^2-(u^3+v^3)z+(u^3v^3)=z^2-12x-64=0$ This has roots 16 and -4 so $u^3=16\rightarrow u=2\sqrt[3]{2}\\v^3=-4\rightarrow v=\sqrt[3]{-4}$ .So $x=u+v=2\sqrt[3]{2}+\sqrt[3]{-4}$ which is $\boxed{0.93}$ to 2 decimal places.

Yes, after many infructuous tries, I put it on wolfram. How are we supposed to know it?