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.

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle dx^3 + ex^2 + fx + g = 0 \ }

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

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x^3 + a x^2 + bx + c = 0 \ }

Identify the three remaining coeficients and use them to calculate values for Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P \ } and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle Q \ } .

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P \ = \frac{3b-a^2} 3 }

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle Q \ = \frac{27c-9ab+2a^3} {27} }

Two transformations change the cubic into a quadratic in Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle w^3 \ } :

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left ( w^3 \right ) ^2 + Q \left ( w^3 \right ) - \frac{P^3}{27} = 0}

Use the values of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P \ } and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle Q \ } to calculate the two values for Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle w^3 \ } .

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle w^3 = \frac{-Q \pm \sqrt{ Q^2 + \frac{4P^3}{27}}} 2}

Often the results will be complex Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle z = x \pm yi \ } . If so change them to polar form.

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle z = r \pm \angle \theta}

Where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r = \sqrt{x^2+y^2}} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \theta = \pm \arctan \frac y x}

Next, find the principle cube root of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle w^3 \ }

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle w_1 = \sqrt[3]{r \pm \angle \ \theta} \ = \ \sqrt[3] r \pm \angle \ \frac{\theta} 3}

And change back to rectangular form:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle z = r \cos \theta + i r \sin \theta \ }

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

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle w_2 = w_1 \left ( - \frac 1 2 + \frac {\sqrt 3} 2 i \right ) }

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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 3Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle w \ } , to calculate the three values of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x \ } .

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x = w - \frac P {3w} - \frac a 3}

An Example with Integer Solutions

Solve the cubic equation.

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x^3 - 4x^2 - 11x + 30 = 0 \ }

Identify the coefficients:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{matrix} a = -4 & b = -11 & c = 30 \end{matrix}}

Substitute these values to get Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P \ } , Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P^3 \ } and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle Q \ } :

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P = \frac{3b - a^2} 3 = \frac{-33-16} 3 = -\frac{49} 3}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P^3 = -\frac{117699}{27}}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle Q = \frac{27c-9ab+2a^3}{27} = \frac{810-369-128}{27} = \frac{286}{27}}

When these are substituted into the quadratic equation:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left ( w^3 \right ) ^2 + Q \left ( w^3 \right ) - \frac{P^3}{27} = 0}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left ( w^3 \right ) ^2 + \frac{286}{27} w^3 + \frac{117699}{27} = 0}

It can be solved by the quadratic formula:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle w^3 = - \frac{143}{27} \pm \frac{20} 3 i \sqrt 3 \approx -5.296296 \pm 11.547005 i }

Changing to polar form:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle w^3 \approx 12.703704 \ \angle \ \pm 2.000839}

The cube root of which is:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle w \approx 2 \frac 1 3 \ \angle \ \pm 0.666946 \approx 1 \frac 5 6 \pm 1.443376 i}

Multiplying by the cube roots of unity:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left (1 \frac 5 6 \pm 1.443376 i \right ) \left ( - \frac 1 2 \pm \frac {\sqrt 3} 2 i \right )}

Gives the other solutions:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle w \approx - \frac 1 3 \pm 2.309401 i}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle w = 2 \frac 1 6 \pm \frac{\sqrt 3} 2 i}

The six solutions

Separating the plus and minus signs gives 6 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle w \ } s:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle w_1 \approx 1 \frac 5 6 + 1.443376 i}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle w_2 \approx 1 \frac 5 6 - 1.443376 i}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle w_3 \approx \frac 1 3 + 2.309401 i}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle w_4 \approx \frac 1 3 - 2.309401 i}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle w_5 = -2 \frac 1 6 + \frac{\sqrt 3} 2 i}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle w_6 = -2 \frac 1 6 - \frac{\sqrt 3} 2 i}

Substituting the 6 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle w \ } s into the formula for $x\$:

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

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

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

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

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

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

$x_{6}=-{\frac {-4}{3}}=-3$

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.