[Math] derivation of the cubic formula

algebra-precalculuscalculuscubics

In my research I came across the following derivation of the cubic formula that was almost complete. However at the end it says to make the substitution $u=z^3$ which would determine the roots. I am not quite sure how to do this and then how to determine the roots and I am looking for some help.

A general form for the cubic equation is,

$$ax^3+bx^2+cx+d=0 \tag{1}$$

To find the roots of this equation we first try to get rid of the quadratic term $x^2$. The substitution $x=y-\dfrac{b}{3a}$ helps in achieving our goal. This results in, $$ay^3+\left(c-\dfrac{b^2}{3a}\right)y+\left(d+\dfrac{2b^3}{27a^2}-\dfrac{bc}{3a}\right)=0\tag{2}$$ which we transform into the following, $$y^3+\dfrac{1}{a}\left(c-\dfrac{b^2}{3a}\right)y+\dfrac{1}{a}\left(d+\dfrac{2b^3}{27a^2}-\dfrac{bc}{3a}\right)=0\tag{3}$$ Upon assuming $e=\dfrac{1}{a}\left(c-\dfrac{b^2}{3a}\right)$ and $f=\dfrac{1}{a}\left(d+\dfrac{2b^3}{27a^2}-\dfrac{bc}{3a}\right)$ we get the equation as, $$y^3+ey+f=0\tag{4}$$ We reduce this equation by the substitution $y=z+\dfrac{s}{z}$ and choosing $s=-\dfrac{e}{3}$ we obtain the simplified equation as, $$z^6+fz^3-\dfrac{e^3}{27}=0\tag{5}$$ What only remains is to make the substitution $u=z^3$.

Best Answer

We start with the substitution $z^3=u$ and obtain a quadratic equation in $u$ \begin{align*} z^6+fz^3-\frac{e^3}{27}&=0\\ u^2+fu-\frac{e^3}{27}&=0\tag{1} \end{align*}

We solve the quadratic equation (1) and get \begin{align*} u_{0,1}=-\frac{f}{2}\pm\sqrt{\frac{f^2}{4}+\frac{e^3}{27}}\tag{2} \end{align*}

Since $z^3=u$ we obtain from (2)

\begin{align*} z_{0,1}=\sqrt[3]{-\frac{f}{2}\pm\sqrt{\frac{f^2}{4}+\frac{e^3}{27}}} \end{align*}

We are free to select either $z_0$ or $z_1$ and we choose $z_0$. Since $z^3=u$ we know there are three solutions \begin{align*} z_0,z_0 e^{\frac{2\pi i}{3}},z_0 e^{\frac{4\pi i}{3}} \tag{3} \end{align*}

From (3) we can calculate the three solutions for $x$ via \begin{align*} y=z-\frac{s}{3z}\qquad\text{ and }\qquad x=y-\frac{b}{3a} \end{align*}

We finally obtain with $\omega_1=e^{\frac{2\pi i}{3}}$ \begin{align*} x_0&=z_0-\frac{s}{3z_0}-\frac{b}{3a}\\ x_1&=z_0\omega_1-\frac{s}{3z\omega_1}-\frac{b}{3a}\\ x_2&=z_0\omega_1^2-\frac{s}{3z\omega_1^2}-\frac{b}{3a} \end{align*}

[2017-11-05] Add-on: Let's make an example to see the formulas in action.

Find the solutions of \begin{align*} \color{blue}{x^3-7x^2+14x-8=0} \end{align*}

  • Step 1: We consider $ax^3+bx^2+cx+d=0$ and calculate $x=y-\frac{b}{3a}$.

    We set $x=y+\frac{7}{3}$ and obtain \begin{align*} y^3-\frac{7}{3}y-\frac{20}{27}=0 \end{align*}

  • Step 2: We consider $y^3+ey+f=0$ and calculate $y=z-\frac{e}{3z}$.

    We set $y=z+\frac{7}{9z}$ which gives \begin{align*} 729z^6-540 z^3+343=0 \end{align*}

  • Step 3: We substitute $z^3=u$ and solve the quadratic equation

    \begin{align*} 729u^2-540 u+343=0 \end{align*}

    We obtain

    \begin{align*} u_{1,2}=\frac{10}{27}\pm\frac{i}{\sqrt{3}} \end{align*}

  • Step 4: The cubic roots $z_0,z_0\omega_1,z_0\omega_2$

    We select $u_1=\frac{10}{27}+\frac{i}{\sqrt{3}}$ and calculate the three cubic roots:

    \begin{align*} z_0&=\sqrt[3]{\frac{10}{27}+\frac{i}{\sqrt{3}}}=\frac{1}{6}\left(5+i\sqrt{3}\right)\\ z_0\omega_1&=z_0e^{\frac{2\pi i}{3}}=\frac{1}{3}\left(-2+i\sqrt{3}\right)\\ z_0\omega_2&=z_0e^{\frac{4\pi i}{3}}=-\frac{1}{6}\left(1+3i\sqrt{3}\right)\\ \end{align*}

  • Step 5: The solutions $x_0,x_1,x_2$

    Finally we can calculate the solutions from $z_0,z_1,z_2$ via \begin{align*} x=y-\frac{b}{3a}=z-\frac{e}{3z}-\frac{b}{3a} \end{align*}

    We obtain \begin{align*} \color{blue}{x_0}&=z_0+\frac{7}{9z_0}+\frac{7}{3}\color{blue}{=4}\\ \color{blue}{x_1}&=z_0\omega_1+\frac{7}{9z_0\omega_1}+\frac{7}{3}\color{blue}{=1}\\ \color{blue}{x_2}&=z_0\omega_2+\frac{7}{9z_0\omega_2}+\frac{7}{3}\color{blue}{=2}\\ \end{align*} and conclude \begin{align*} \color{blue}{x^3-7x^2+14x-8=(x-1)(x-2)(x-4)=0} \end{align*}

Note: Working with the conjugate complex $u_2=\overline{u_1}=\frac{10}{27}-\frac{i}{\sqrt{3}}$ instead of $u_1$ gives the complex conjugates $\overline{z_1},\overline{z_1}e^{-\frac{2\pi i}{3}}$ and $\overline{z_1}e^{-\frac{4\pi i}{3}}$ and leads (of course) to the same results $x_0=4,x_1=1$ and $x_2=2$.

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