Can $x^3+3x^2+1=0$ be solved using high school methods? I encountered the following problem in a high-school math text, which I wasn't able to solve using factorization/factor theorem:
Solve $x^3+3x^2+1=0$
Am I missing something here, or is indeed a more advanced method necessary to solve this particular cubic? The answer provided was $x\doteq-3.1$, which I was only able to confirm using CAS.

You might be able to get away with using some "miracle" substitutions that solve the cubic (which can be generalised to solve a general cubic).
Begin with
$$x^3+3x^2+1=0 \tag{$\dagger$}$$


*

*Substitute $x = y - 1$ into $(\dagger)$ to obtain
$$y^3 - 3y + 3 = 0. \tag{$\star$}$$

*Let $y = z + \frac{1}{z}$ so that $(\star)$ becomes $z^3+\frac{1}{z^3}+3 = 0$ or by multiplying through by $z^3$,
$$\left(z^3\right)^2 + 3\left(z^3\right) + 1 = 0. \tag{$\ast$}$$

*Solve $(\ast)$ for $z^3$ with the quadratic equation formula,
$$z^3 = \frac{-3\pm\sqrt{5}}{2}.$$
The two real solutions of $(\ast)$ are thus $z_1 = -\sqrt[3]{(3+\sqrt{5})/2}$ and $z_2 = -\sqrt[3]{(3-\sqrt{5})/2}$.

*Unravel the substitutions: $x = y - 1 = z + \frac{1}{z} - 1$ so solutions to $(\dagger)$ are
$$\begin{align*}x_1&= z_1 + 1/z_1 - 1 = -\sqrt[3]{(3+\sqrt{5})/2} - \sqrt[3]{2/(3+\sqrt{5})} - 1\\
    x_2&= z_2 + 1/z_2 - 1 =-\sqrt[3]{(3-\sqrt{5})/2} - \sqrt[3]{2/(3-\sqrt{5})} - 1\end{align*}$$
It actually turns out that $x_1 = x_2$ (another miracle!). So we may write the final solution in exact form as
$$x = -\sqrt[3]{(3+\sqrt{5})/2} - \sqrt[3]{2/(3+\sqrt{5})} - 1 \approx -3.1$$
