Cubic equation/Proofs

Vieta and Harriot developed a plan for solving cubic equations that involves transforming the original equation into a depressed cubic - one without a quadratic term. They introduced a second substitution that transforms it 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.

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 mathematicians Vieta and Harriot 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 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.

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\$:

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

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

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

Substitute these into the quadratic equation in $w^{2}\$:

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

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

The Quadratic Formula Gives:

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

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:

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

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

Separating the plus and minus signs gives 6 $w\$s:

$w_{1}\approx 1{\frac {5}{6}}+1.443376i$

$w_{2}\approx 1{\frac {5}{6}}-1.443376i$

$w_{3}\approx {\frac {1}{3}}+2.309401i$

$w_{4}\approx {\frac {1}{3}}-2.309401i$

$w_{5}=-2{\frac {1}{6}}+{\frac {\sqrt {3}}{2}}i$

$w_{6}=-2{\frac {1}{6}}-{\frac {\sqrt {3}}{2}}i$

Substituting the 6 $w\$s into Vieta's formula for $t\$:

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

$t_{1}\approx 1{\frac {5}{6}}+1.443376i-{\frac {\frac {-49}{3}}{3(1{\frac {5}{6}}+1.443376i)}}\ =\ 3{\frac {2}{3}}$

$t_{2}\approx 1{\frac {5}{6}}-1.443376i-{\frac {\frac {-49}{3}}{3(1{\frac {5}{6}}-1.443376i)}}\ =\ 3{\frac {2}{3}}$

$t_{3}\approx -{\frac {1}{3}}+2.309401i-{\frac {\frac {-49}{3}}{3(-{\frac {1}{3}}+2.309401i)}}\ =\ {\frac {2}{3}}$

$t_{4}\approx -{\frac {1}{3}}-2.309401i-{\frac {\frac {-49}{3}}{3(-{\frac {1}{3}}-2.309401i)}}\ =\ {\frac {2}{3}}$

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

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

And finally substituting the 3 $t\$s into the formula for $x\$ that gave the depressed cubic:

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

$x_{1}=3{\frac {2}{3}}-{\frac {-4}{3}}=5$

$x_{2}={\frac {2}{3}}-{\frac {-4}{3}}=2$

$x_{3}=-4{\frac {1}{3}}-{\frac {-4}{3}}=-3$

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