Cubic Equations

Cardano’s formula:

  1. Given {\displaystyle x^{3}+a x^{2}+b x+c=0}
  2. Use shifts {\displaystyle x=y+t} to eliminate the {y^2} term.

    { (y+t)^{3}+a (y+t)^{2}+b (y+t)+c=0} doesn’t have {y^2} term if {3t+a =0.} Therefore {x=y-\frac{a}{3}.}
  3. The equation reduces to the depressed cubic {\displaystyle y^3+py+q=0.}
  4. Substitute {\displaystyle y=u+v}
    The equation becomes {\displaystyle (u+v)^3 +3p(u+v) +q = u^{3}+v^{3}+q+3uv(u+v)+p(u+v).}
  5. We want {u, v} such that { u^{3}+v^{3}+q=0, 3uv+p=0.}
  6. {u^3, v^3} solve the quadratic equation {z^{2}-\left(u^{3}+v^{3}\right) z+(u v)^{3}=z^{2}+q z-\frac{p^{3}}{27}= 0}
  7. {\displaystyle u=\sqrt[3]{-\frac{q}{2}+\sqrt{\frac{q^{2}}{4}+\frac{p^{3}}{27}}}, \quad \quad v=\sqrt[3]{-\frac{q}{2}-\sqrt{\frac{q^{2}}{4}+\frac{p^{3}}{27}}} }
  8. {\displaystyle y =u+v =\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}}}}
  9. {\displaystyle x=y-\frac{a}{3}= -\frac{a}{3}+ \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}}}}
  10. To get all the roots, once has to choose the cube roots properly. The two choices of cube roots used above have to be conjugates to each other.

Vieta’s Substitution:

  1. Given {\displaystyle x^{3}+a x^{2}+b x+c=0}, reduced to the depressed cubic {\displaystyle y^3+py+q=0.}
  2. Now substitute {\displaystyle y = z-\frac{p}{3 z}.}
  3. The equation transforms to {\displaystyle z^{3}-\frac{p^{3}}{27 z^{3}}+q=0}, that is {\displaystyle z^{6}+q\left(z^{3}\right)-\frac{p^{3}}{27}=0}
  4. {\displaystyle z^{6}+q\left(z^{3}\right)-\frac{p^{3}}{27}=0} is quadratic equation in {z^3.}
  5. {\displaystyle z^3=-\frac{q}{2} \pm \sqrt{\frac{p^{3}}{27}+\frac{q^{2}}{4}.}}
  6. { y =z-\frac{p}{3 z}} for {z = \sqrt[3]{-\frac{q}{2} \pm \sqrt{\frac{p^{3}}{27}+\frac{q^{2}}{4}}}}. For different choices of the cube roots we get different roots {y.}

Trigonometric method:

  1. Given {\displaystyle x^{3}+a x^{2}+b x+c=0}, reduced to the depressed cubic {\displaystyle y^3+py+q=0.}
  2. Substitute {y =a\cos\theta.}
  3. we want to make {\displaystyle y^3+py+q =0} equivalent to {4 \cos ^{3} \theta-3 \cos \theta -A}
  4. {a=2 \sqrt{-\frac{p}{3}}} gives {\displaystyle 4 \cos ^{3} \theta-3 \cos \theta-\frac{3 q}{2 p} \sqrt{\frac{-3}{p}}=0.}
  5. Therefore using {\displaystyle \cos (3\theta) =4 \cos ^{3} \theta-3 \cos \theta}, we get {\cos( 3\theta) =\frac{3 q}{2 p} \sqrt{\frac{-3}{p}}.}
  6. {\displaystyle \theta= \frac{1}{3} \arccos \left(\frac{3 q}{2 p} \sqrt{\frac{-3}{p}}\right) + \frac{2\pi m}{3}}
  7. {y=\displaystyle2 \sqrt{-\frac{p}{3}} \cos \left(\frac{1}{3} \arccos \left(\frac{3 q}{2 p} \sqrt{\frac{-3}{p}}\right)-\frac{2 \pi m}{3}\right)} are the roots of the depressed cubic.

Lagrange Resolvents:

  1. Let {x_1, x_2, x_3} be the roots of the cubic equation {\displaystyle x^{3}+a x^{2}+b x+c=0.} Consider the quantities (Fourier transform)

    \displaystyle \begin{aligned} y_0 =&x_1+x_2+x_3\\ y_1 =& x_{1}+\omega x_{2}+\omega^{2} x_{3}\\ y_2 =& x_{1}+\omega^2 x_{2}+\omega x_{3} \end{aligned}


  2. \displaystyle \begin{aligned} y_0^3+y_1^3 +y_2^3 =& \left (x_{1}+x_{2}+x_{3} \right)^3+\left (x_{1}+\omega x_{2}+\omega^{2} x_{3} \right)^3 + \left (x_{1}+\omega^2 x_{2}+\omega x_{3} \right)^3\\ =&2\left(x_{1}^{3}+x_{2}^{3}+x_{3}^{3}\right)-3\left(x_{1}^{2} x_{2}+x_{1} x_{2}^{2}+x_{1}^{2} x_{3}+x_{1} x_{3}^{2}+x_{2}^{2} x_{3}+x_{2} x_{3}^{2}\right) +12x_1x_2x_3\\ \end{aligned}

    is a symmetric as a polynomial in {x_1, x_2, x_3} and hence rational.
  3. \displaystyle \begin{aligned} y_1y_2 =x_1^2+x_2^2+x_3^2-x_1x_2-x_2x_3-x_1x_3 \end{aligned}
    is also symmetric and rational.
  4. Therefore {y_1^3,y_2^3} are roots of a quadratic equation with known coefficients (write all these symmetric polynomials in terms of elementary symmetric polynomials)
  5. Solve the quadratic and use the value of {y_0, y_1, y_2} to compute {x_1, x_2, x_3.} (Use Inverse of the above transform)

Discriminant, repeated roots

\displaystyle f(x)=a x^{3}+b x^{2}+c x+d

The discriminant of the cubic is

\displaystyle \Delta=18 a b c d-4 b^{3} d+b^{2} c^{2}-4 a c^{3}-27 a^{2} d^{2}

The sign of {\Delta} tells if you have three real roots or just one real root.

{\Delta>0}, then we have three distinct real roots.
{\Delta<0}, we have one real roots and two conjugate complex roots.
{\Delta=0}, we have three real roots, with one of them repeating twice.

\displaystyle \delta=\sqrt{\frac{b^{2}-3 a c}{9 a^{2}}}

Depending on the sign of {b^2-3ac,} the cubic {f(x)} will have either two local extrema or no local extrema.
{b^2-3ac >0} implies that there are two local extrema.
{b^2-3ac <0} implies there are no local extrema.
{b^2-3ac=0} corresponds to one stationary saddle point.

TARTAGLIA’S ORIGINAL POEM

Quando chel cubo con le cose appresso
Se aqquaglia ?a qualche numero discreto
Trouan duo altri differenti in esso
Dapoi terrai questo per consueto
Che?llor productto sempre sia equale
Alterzo cubo delle cose neto,
El residuo poi suo generale
Delli lor lati cubi ben sottrati
Varra la tua cosa principale.
In el secondo de cotestiatti
Quando chel cubo restasse lui solo
Tu osseruarai questaltri contratti,
Del numer farai due tal part?`a uolo
Che luna in laltra si produca schietto
El terzo cubo delle cose in stolo
Delle qual poi, per communprecetto
Torrai li lati cubi insieme gionti
Et cotal somma sara il tuo concetto.
El terzo poi de questi nostri conti
Se solue col secondo se ben guardi
Che per natura son quasi congionti.
Questi trouai, non con passi tardi
Nel mille cinquecent`e, quatroe trenta
Con fondamenti ben sald`e gagliardi
Nella citta dal marintorno centa.

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