Formula for the Roots of a Cubic Equation
Formulas
The solution of the cubic equation $t^{3}+pt+q = 0$ is as follows.
$$ \begin{cases} t_{1}=u_{1}+v_{1}=\sqrt[3]{-\frac{q}{2}+\sqrt{\frac{q^{2}}{4}+\frac{p^{3}}{27}}}+\sqrt[3]{-\frac{q}{2}-\sqrt{\frac{q^{2}}{4}+\frac{p^{3}}{27}}} \\ t_{2}=u_{2}+v_{3}=\sqrt[3]{-\frac{q}{2}+\sqrt{\frac{q^{2}}{4}+\frac{p^{3}}{27}}}\omega+\sqrt[3]{-\frac{q}{2}-\sqrt{\frac{q^{2}}{4}+\frac{p^{3}}{27}}}\omega^{2} \\ t_{3}=u_{3}+v_{2}=\sqrt[3]{-\frac{q}{2}+\sqrt{\frac{q^{2}}{4}+\frac{p^{3}}{27}}}\omega^{2}+\sqrt[3]{-\frac{q}{2}-\sqrt{\frac{q^{2}}{4}+\frac{p^{3}}{27}}}\omega\end{cases} $$
Here $\omega = e^{i\frac{2}{3}\pi}$.
Proof
Cardano’s Method
Let us consider a cubic equation $ax^{3}+bx^{2}+cx+d=0(a\ne0)$ is given. To simplify the solution, without loss of generality, it is shown as follows.
$$ \begin{equation} x^{3}+ax^{2} +bx+c=0 \end{equation} $$
To eliminate the quadratic term, substitute $x=t-{\textstyle \frac{a}{3}}$. Then it becomes:
$$ \begin{align*} &&\left( t- \frac{a}{3}\right)^{3}+a\left( t-\frac{a}{3} \right)^{2}+b\left( t-\frac{a}{3} \right) + c &=0 \\ \implies && \left( t^{3}-\cancel{at^{2}}+\frac{a^{2}}{3}t-\frac{a^{3}}{27} \right)+\left(\cancel{at^{2}}-\frac{2a^{2}}{3}t + \frac{a^{3}}{9}\right) + \left( bt-\frac{ab}{3} \right) +c &= 0 \\ \implies && t^{3}+\left( \frac{a^{2}}{3} -\frac{2a^{2}}{3}+b \right)t +\left( -\frac{a^{3}}{27}+\frac{a^{3}}{9}-\frac{ab}{3}+c \right)&=0 \\ \implies && t^{3}+\left( b- \frac{a^{2}}{3} \right)t +\left( \frac{2a^{3}}{27}-\frac{ab}{3}+c \right)&=0 \end{align*} $$
To further simplify the equation, let’s set $p=b-\frac{a^{2}}{3}$, $q=\frac{2a^{3}}{27}-\frac{ab}{3}+c$; then the above equation becomes as follows.
$$ \begin{equation} t^{3}+pt+q = 0 \end{equation} $$
If we substitute once more with $t=u+v$, the above equation becomes as follows.
$$ \begin{equation} (u+v)^{3}+p(u+v)+q=0 \end{equation} $$
Also, according to the multiplication formula, the following equation holds.
$$ (u+v)^{3}=u^{3}+3u^{2}v+3uv^{2}+v^{2}=u^{3}+v^{3}+3uv(u+v) $$
Transferring the entire right-hand side to the left yields the equation below.
$$ \begin{equation} (u+v)^{3} - 3uv(u+v) - (u^{3}+v^{3})=0 \end{equation} $$
Comparing $(3)$ and $(4)$, it can be seen that solving $(3)$ is equivalent to finding $u$, $v$ that satisfy $p=-3uv$, $q=-(u^{3}+v^{3})$. Writing both equations for $u$, $v$ gives us:
$$ \begin{align*} u^{3}v^{3} &= -\frac{p^{3}}{27} \\ u^{3}+v^{3} &= -q \end{align*} $$
Considering Vieta’s formulas for quadratic equations, we can see that $u^{3}$, $v^{3}$ are the roots of the following quadratic equation.
$$ X^{2} +qX -\frac{p^{3}}{27}=0 $$
Then, the quadratic formula yields the following equation.
$$ \begin{equation} \begin{aligned} X_{1}&=u^{3}=\frac{-q+\sqrt{q^{2}+\frac{4p^{3}}{27}}}{2}=-\frac{q}{2}+\sqrt{\frac{q^{2}}{4}+\frac{p^{3}}{27}} \\ X_{2}&= v^{3}=\frac{-q-\sqrt{q^{2}+\frac{4p^{3}}{27}}}{2}=-\frac{q}{2}-\sqrt{\frac{q^{2}}{4}+\frac{p^{3}}{27}} \end{aligned} \end{equation} $$
At this point, for any real number $\alpha$, the three imaginary roots that satisfy $z^{3}=\alpha$ are as follows.
$$ z_{1}=\sqrt[3]{\alpha} \quad \text{and} \quad z_{2}=\sqrt[3]{\alpha}\omega \quad \text{and} \quad z_{3}=\sqrt[3]{\alpha}\omega^{2} $$
Here $\omega$ is a complex number satisfying $\omega^{3}=1$, specifically $\omega=e^{i\frac{2}{3}\pi}=\cos (\frac{2}{3}\pi)+i\sin (\frac{2}{3}\pi)$. Therefore, $u, v$ that satisfies $(5)$ is as follows.
$$ \begin{cases} u_{1}=\sqrt[3]{-\frac{q}{2}+\sqrt{\frac{q^{2}}{4}+\frac{p^{3}}{27}}} \\ u_{2}=\sqrt[3]{-\frac{q}{2}+\sqrt{\frac{q^{2}}{4}+\frac{p^{3}}{27}}}\omega \\ u_{3}=\sqrt[3]{-\frac{q}{2}+\sqrt{\frac{q^{2}}{4}+\frac{p^{3}}{27}}}\omega^{2}\end{cases}\quad \text{and} \quad \begin{cases} v_{1}=\sqrt[3]{-\frac{q}{2}-\sqrt{\frac{q^{2}}{4}+\frac{p^{3}}{27}}} \\ v_{2}=\sqrt[3]{-\frac{q}{2}-\sqrt{\frac{q^{2}}{4}+\frac{p^{3}}{27}}}\omega \\ v_{3}=\sqrt[3]{-\frac{q}{2}-\sqrt{\frac{q^{2}}{4}+\frac{p^{3}}{27}}}\omega^{2}\end{cases} $$
However, since $p=-uv$ is a real number, only the combination that results in a real number when multiplied is the solution. Thus, the solution is as follows.
$$ \begin{cases} t_{1}=u_{1}+v_{1}=\sqrt[3]{-\frac{q}{2}+\sqrt{\frac{q^{2}}{4}+\frac{p^{3}}{27}}}+\sqrt[3]{-\frac{q}{2}-\sqrt{\frac{q^{2}}{4}+\frac{p^{3}}{27}}} \\ t_{2}=u_{2}+v_{3}=\sqrt[3]{-\frac{q}{2}+\sqrt{\frac{q^{2}}{4}+\frac{p^{3}}{27}}}\omega+\sqrt[3]{-\frac{q}{2}-\sqrt{\frac{q^{2}}{4}+\frac{p^{3}}{27}}}\omega^{2} \\ t_{3}=u_{3}+v_{2}=\sqrt[3]{-\frac{q}{2}+\sqrt{\frac{q^{2}}{4}+\frac{p^{3}}{27}}}\omega^{2}+\sqrt[3]{-\frac{q}{2}-\sqrt{\frac{q^{2}}{4}+\frac{p^{3}}{27}}}\omega\end{cases} $$
Since all cubic equations can be represented without a quadratic term as $(2)$ through substitution, the above formula is sufficient. If it’s expressed as a formula for $(1)$, it is as follows.
$$ \begin{cases} x_{1}=\sqrt[3]{-\frac{1}{2}\left(\frac{2a^{3}}{27}-\frac{ab}{3}+c \right)+\sqrt{\frac{1}{4}\left( \frac{2a^{3}}{27}-\frac{ab}{3}+c \right)^{2}+\frac{1}{27}\left(b-\frac{a^{2}}{3} \right)^{3}}}+\sqrt[3]{-\frac{1}{2}\left(\frac{2a^{3}}{27}-\frac{ab}{3}+c \right)-\sqrt{\frac{1}{4}\left( \frac{2a^{3}}{27}-\frac{ab}{3}+c \right)^{2}+\frac{1}{27}\left(b-\frac{a^{2}}{3} \right)^{3}}}-\frac{a}{3} \\ x_{2}=\sqrt[3]{-\frac{1}{2}\left(\frac{2a^{3}}{27}-\frac{ab}{3}+c \right)+\sqrt{\frac{1}{4}\left( \frac{2a^{3}}{27}-\frac{ab}{3}+c \right)^{2}+\frac{1}{27}\left(b-\frac{a^{2}}{3} \right)^{3}}}\omega+\sqrt[3]{-\frac{1}{2}\left(\frac{2a^{3}}{27}-\frac{ab}{3}+c \right)-\sqrt{\frac{1}{4}\left( \frac{2a^{3}}{27}-\frac{ab}{3}+c \right)^{2}+\frac{1}{27}\left(b-\frac{a^{2}}{3} \right)^{3}}}\omega^{2} -\frac{a }{3} \\ x_{3}=\sqrt[3]{-\frac{1}{2}\left(\frac{2a^{3}}{27}-\frac{ab}{3}+c \right)+\sqrt{\frac{1}{4}\left( \frac{2a^{3}}{27}-\frac{ab}{3}+c \right)^{2}+\frac{1}{27}\left(b-\frac{a^{2}}{3} \right)^{3}}}\omega^{2}+\sqrt[3]{-\frac{1}{2}\left(\frac{2a^{3}}{27}-\frac{ab}{3}+c \right)-\sqrt{\frac{1}{4}\left( \frac{2a^{3}}{27}-\frac{ab}{3}+c \right)^{2}+\frac{1}{27}\left(b-\frac{a^{2}}{3} \right)^{3}}}\omega -\frac{a }{3} \end{cases} $$
■