Is there anything like “cubic formula”?
Solution 1:
Yes, we do have a cubic formula! By Cardan's Method...
Cardan's Method: To solve the general cubic$$x^3+ax^2+bx+c=0\tag{i}$$ Remove the $ax^2$ term by substituting $x=\dfrac {y-a}3$. Let the transformed equation be$$y^3+qy+r=0\tag{ii}$$ To solve this depressed cubic, substitute $y=u+v$ to get$$u^3+v^3+(3uv+q)(u+v)+r=0\tag{iii}$$ Put $3uv+q=0$ to get $u=-\dfrac q{3v}$ and substituting this back gives a quadratic in $v^3$. The roots of the quadratic are equal to $u^3,v^3$ respectively. And from our substitution, we get a root as$$y=\left\{-\frac r2+\sqrt{\frac {r^2}4+\frac {q^3}{27}}\right\}^{\frac 13}+\left\{-\frac r2-\sqrt{\frac {r^2}4+\frac {q^3}{27}}\right\}^{\frac 13}\tag{iv}$$ With the other two roots found with the cube roots of unity.
To find the original root of $(\text i)$, substitute $y$ into your transformation.
Solution 2:
One method is to depress the cubic, then apply trig functions.
$$0=sx^3+tx^2+ux+v$$
Divide both sides by $s$ to get:
$$0=x^3+ax^2+bx+c$$
Let $x=y-\frac a3$ to get
$$0=y^3+\underbrace{\left(b-\frac{a^2}3\right)}_dy+\underbrace{c-\frac{ab}3+\frac{2a^3}{27}}_e=y^3+dy+e$$
If $d>0$, then use the trigonometric identity:
$$\sinh(3\theta)=4\sinh^3(\theta)+3\sinh(\theta)$$
where
$$\sinh(\theta)=\frac{e^\theta-e^{-\theta}}2$$
We exploit this identity by letting $y=fz$ and multiplying both sides by $g$ to get
$$0=f^3gz^3+dfgz+eg$$
$$\begin{cases}4=f^3g\\3=dfg\end{cases}\implies\begin{cases}f=2\sqrt{\frac d3}\\g=\frac{3\sqrt3}{2d^{3/2}}\end{cases}$$
$$0=4z^3+3z+\frac{3e\sqrt3}{2d^{3/2}}=\sinh(3\operatorname{arcsinh}(z))+\frac{3e\sqrt3}{2d^{3/2}}$$
$$\implies\sinh(3\operatorname{arcsinh}(z))=-\frac{3e\sqrt3}{2d^{3/2}}$$
$$\implies z=-\sinh\left(\frac13\operatorname{arcsinh}\left(\frac{3e\sqrt3}{2d^{3/2}}\right)\right)$$
$$\implies x=-2\sqrt{\frac d3}\sinh\left(\frac13\operatorname{arcsinh}\left(\frac{3e\sqrt3}{2d^{3/2}}\right)\right)-\frac a3$$
If $d<0$, use $\cos(3\theta)$ or $\cosh(3\theta)$ and respective triple angle formulas.