Cubic equation/Proofs

Vieta and Harriot helped to develop a plan for solving cubic equations that involves transforming the original equation into a depressed cubic - one without a quadratic term. They have a second substitution that transforms the cubic into a sixth degree equation which is of quadratic form - having only terms of degrees six, three and zero. This can be solved by the quadratic formula and then taking the cube root. The quadratic formula often results in complex solutions. Finding the cube root of a complex number can be accomplished in polar form, so changing the form of the complex numbers as needed, is appropriate. After the three cube roots are found, the values must be substituted back through the two transformations, to give the solutions of the original equation.

Using the method

To solve a cubic equation with this method, collect all the terms in decreasing degree on one side of the equation.

$dx^{3}+ex^{2}+fx+g=0\$

If the coefficient of the cubic term is not 1, divide the equation by that coefient.

$x^{3}+ax^{2}+bx+c=0\$

Identify the three remaining coeficients and use them to calculate values for $P\$ and $Q\$.

$P\ ={\frac {3b-a^{2}}{3}}$

$Q\ ={\frac {27c-9ab+2a^{3}}{27}}$

Two transformations change the cubic into a quadratic in $w^{3}\$:

$\left(w^{3}\right)^{2}+Q\left(w^{3}\right)-{\frac {P^{3}}{27}}=0$

Use the values of $P\$ and $Q\$ to calculate the two values for $w^{3}\$.

$w^{3}={\frac {-Q\pm {\sqrt {Q^{2}+{\frac {4P^{3}}{27}}}}}{2}}$

Often the results will be complex $z=x\pm yi\$. If so change them to polar form.

$z=r\pm \angle \theta$

Where $r={\sqrt {x^{2}+y^{2}}}$ and $\theta =\pm \arctan {\frac {y}{x}}$

Next, find the principle cube root of $w^{3}\$

$w_{1}={\sqrt[{3}]{r\pm \angle \ \theta }}\ =\ {\sqrt[{3}]{r}}\pm \angle \ {\frac {\theta }{3}}$

And change back to rectangular form:

$z=r\cos \theta +ir\sin \theta \$

The other two roots are found by multiplying by the cube roots of unity.

$w_{2}=w_{1}\left(-{\frac {1}{2}}+{\frac {\sqrt {3}}{2}}i\right)$

$w_{3}=w_{1}\left(-{\frac {1}{2}}-{\frac {\sqrt {3}}{2}}i\right)$

The formula below is a combination of the two transformations that were use to get the quadratic equation solved above. Due to the plus-and-minus in the 3$w\$s from the quadratic formula, there are often six different results at this point. But in the final step, there will be only three unique answers.

Use each of the six values of $w\$, to calculate the three values of $x\$.

$x=w-{\frac {P}{3w}}-{\frac {a}{3}}$

An Example with Integer Solutions

Solve the cubic equation.

$x^{3}-4x^{2}-11x+30=0\$

Identify the coefficients:

${\begin{matrix}a=-4&b=-11&c=30\end{matrix}}$

Substitute these values to get $P\$, $P^{3}\$ and $Q\$:

$P={\frac {3b-a^{2}}{3}}={\frac {-33-16}{3}}=-{\frac {49}{3}}$

$P^{3}=-{\frac {117699}{27}}$

$Q={\frac {27c-9ab+2a^{3}}{27}}={\frac {810-369-128}{27}}={\frac {286}{27}}$

When these are substituted into the quadratic equation:

$\left(w^{3}\right)^{2}+Q\left(w^{3}\right)-{\frac {P^{3}}{27}}=0$

$\left(w^{3}\right)^{2}+{\frac {286}{27}}w^{3}+{\frac {117699}{27}}=0$

It can be solved by the quadratic formula:

$w^{3}={\frac {-\left({\frac {286}{27}}\right)\pm {\sqrt {\left({\frac {286}{27}}\right)^{2}-4\left({\frac {117699}{27}}\right)}}}{2}}$

Which simplifies to:

$w^{3}=-{\frac {143}{27}}\pm {\frac {20}{3}}i{\sqrt {3}}\approx -5.296296\pm 11.547005i$

Changing to polar form:

$w^{3}\approx 12.703704\ \angle \ \pm 2.000839$

The cube root of which is:

$w\approx 2{\frac {1}{3}}\ \angle \ \pm 0.666946\approx 1{\frac {5}{6}}\pm 1.443376i$

Multiplying by the cube roots of unity:

$\left(1{\frac {5}{6}}\pm 1.443376i\right)\left(-{\frac {1}{2}}\pm {\frac {\sqrt {3}}{2}}i\right)$

Gives the other solutions:

$w\approx -{\frac {1}{3}}\pm 2.309401i$

$w=2{\frac {1}{6}}\pm {\frac {\sqrt {3}}{2}}i$

Substituting the 6 $w\$s into the formula for $x\$:

$x=w-{\frac {P}{3w}}-{\frac {a}{3}}$

$x_{1}=1{\frac {5}{6}}+1.443376i-{\frac {\frac {-49}{3}}{1{\frac {5}{6}}+1.443376i}}-{\frac {-4}{3}}=5$

$x_{2}=1{\frac {5}{6}}-1.443376i-{\frac {\frac {-49}{3}}{1{\frac {5}{6}}-1.443376i}}-{\frac {-4}{3}}=5$

$x_{3}={\frac {1}{3}}+2.309401i-{\frac {\frac {-49}{3}}{{\frac {1}{3}}+2.309401i}}-{\frac {-4}{3}}=2$

$x_{4}={\frac {1}{3}}-2.309401i-{\frac {\frac {-49}{3}}{{\frac {1}{3}}-2.309401i}}-{\frac {-4}{3}}=2$

$x_{5}=-2{\frac {1}{6}}+{\frac {\sqrt {3}}{2}}i-{\frac {\frac {-49}{3}}{-2{\frac {1}{6}}+{\frac {\sqrt {3}}{2}}i}}-{\frac {-4}{3}}=-3$

$x_{6}=-2{\frac {1}{6}}-{\frac {\sqrt {3}}{2}}i-{\frac {\frac {-49}{3}}{-2{\frac {1}{6}}-{\frac {\sqrt {3}}{2}}i}}-{\frac {-4}{3}}=-3$

Which can be checked to be the three solutions to the original equation.

The Derivation of the Method

The Depressed Cubic

Start with a cubic equation in this form:

$x^{3}+ax^{2}+bx+c=0\$

Let $x=t-{\frac {a}{3}}\ \$

Substituting this for $x\$ in the original equation gives::

$\left(t-{\frac {a}{3}}\right)^{3}+a\left(t-{\frac {a}{3}}\right)^{2}+b\left(t-{\frac {a}{3}}\right)+c=0$

Expanding:

$t^{3}-at^{2}+{\frac {a^{2}t}{3}}-{\frac {a^{3}}{27}}+at^{2}-{\frac {2a^{2}t}{3}}+{\frac {a^{3}}{9}}+bt-{\frac {ab}{3}}+c=0$

Gathering like terms in $t\$:

$t^{3}+(at^{2}-at^{2})+\left({\frac {a^{2}t}{3}}-{\frac {2a^{2}t}{3}}+by\right)+\left(-{\frac {a^{3}}{27}}+{\frac {a^{3}}{9}}-{\frac {ab}{3}}+c\right)$

Simplifying by adding like terms causes the quadratic terms to drop out:

$t^{3}+\left({\frac {3b-a^{2}}{3}}\right)t+\left({\frac {27c-9ab+2a^{3}}{27}}\right)=0$

Common practice uses $P\$ for ${\frac {3b-a^{2}}{3}}$

and $Q\$ for ${\frac {27c-9ab+2a^{3}}{27}}$

Gives what is called the depressed cubic:

$t^{3}+Pt+Q=0\$

Vieta's Substitution

The mathematician Vieta devised a plan that allows the depressed cubic to be written as a sixth degree equation, which is in the form of a quadratic.

Start with:

$t=w-{\frac {P}{3w}}$

Substituting;

$\left(w-{\frac {P}{3w}}\right)^{3}+P\left(w-{\frac {P}{3w}}\right)+Q=0$

Expanding:

$w^{3}-3w^{2}\left({\frac {P}{3w}}\right)+3w\left({\frac {P^{2}}{9w^{2}}}\right)-\left({\frac {P^{3}}{27w^{3}}}\right)+Pw-{\frac {P^{2}}{3w}}+Q=0$

Simplifying:

$w^{3}-wP+{\frac {P^{2}}{3w}}-\left({\frac {P^{3}}{27w^{3}}}\right)+wP-{\frac {P^{2}}{3w}}+Q=0$

The second term and fifth term cancel, as do the third and sixth; leaving:

$w^{3}-\left({\frac {P^{3}}{27w^{3}}}\right)+Q=0$

Multiplying the equation by $w^{3}\$ yields a sixth degree equation:

$w^{6}+Qw^{3}-{\frac {P^{3}}{27}}=0$

Which can be written as a quadratic in $w^{3}\$.

$\left(w^{3}\right)^{2}+Q\left(w^{3}\right)-{\frac {P^{3}}{27}}=0$

Finishing the Solution

Using the quadratic formula to solve for $w^{3}\$

$w^{3}={\frac {-Q\pm {\sqrt {Q^{2}+{\frac {4P^{3}}{27}}}}}{2}}$

Finding the principle cube root:

$w_{1}={\sqrt[{3}]{\frac {-Q\pm {\sqrt {Q^{2}+{\frac {4P^{3}}{27}}}}}{2}}}$

And the other two roots are found by multiplying by a cube root of unity.

$w_{2}=w_{1}\left(-{\frac {1}{2}}+{\frac {\sqrt {3}}{2}}i\right)$

$w_{3}=w_{1}\left(-{\frac {1}{2}}-{\frac {\sqrt {3}}{2}}i\right)$

Since the cube root has a plus-and-minus sign in it, there are six solutions for $w\$. But as they are substituted into Vieta's formula to get $t\$, the result is three pairs of identical solutions.

Finally, the 3 $t\$s are substituted into the original formula for $x\$, to get the three solutions of the cubic equation.