What are the ways to solve cubic equations?

Question

Cubic equations of the form $ax^3+bx^2+cx+d$ can be solved in various ways. Some are easy to easy to factor in a pair, for some the roots can be found out by trial-and-error, some are one-of-a-kind, some can be reduced to a quadratic equation. A compilation of all possible ways to solve cubic equations would be very helpful for students and learners.

Answer 1

One-of-a-kind cubic equations

1. $x^3-6x^2+12x-8=0 $

2. $\implies x^3-2^3-6x^2+12x=0$

3. $a^3-b^3=(a-b)[(a-b)^2+3ab]\\\therefore x^3-2^3=(x-2)[(x-2)^2+3\times x\times2]$

4. $6x^2+12x=6x(x-2)$

5. $\implies (x-2)[(x-2)^2+6x]-6x(x-2)=0$

6. $\implies(x-2)[(x-2)^2+6x-6x]=0$

7. $\implies (x-2)(x-2)^2=0\\ \implies(x-2)^3=0 \\ \implies x-2=0$

8. $\implies \boxed{x=2}$

Answer 2

As with quadratics, first we divide through to make the polynomial monic and then shift the unknown by a constant so the second highest power has zero coefficient, giving $x^3+px+q=0$. We now use Cardano's method. There exist complex numbers $u,\,v$ with $u+v=x,\,uv=-\frac{p}{3}$ (since if only you knew $x$ we'd just have to solve $t^2-xt-\frac{p}{3}=0$) so $u^3+v^3=x(x^2+p)=-q$ and $u^3v^3=-\frac{p^3}{27}$. Solving a quadratic gives $u^3,\,v^3$, so $u=u_0\omega^n,\,v=v_0\omega^{-n}$ say with $\omega=\exp\frac{2\pi i}{3},\,n\in\{ 0,\,1,\,2\}$. Summing gives three values for $x=u+v$.

Answer 3

$ax^3+bx^2+cx+d=0$

1. Divide to a to get $x^3$ term 1

$$x^3+\frac{b}{a}x^2+\frac{c}{a} x+\frac{d}{a}=0$$

$$x^3+b_1 x^2+c_1 x+d_1=0$$

2. Eliminate $x^2$ term via using $x=z-\frac{b_1}{3}$ transform Then you will get

$$(z-\frac{b_1}{3})^3+b_1(z-\frac{b_1}{3})^2+c_1(z-\frac{b_1}{3})+d_1=0$$

$$z^3+c_2 z+d_2=0$$

3. Use binom expansion $$(p+y)^3=p^3+3p^2y+3py^2+y^3$$ Reorder the equation as: $$(p+y)^3-3py(p+y)-(p^3+y^3)=0$$

Define: $p+y=z$ Thus

$$3.p.y=-c_2$$
$$p^3+y^3=-d_2$$

$$p=-\frac{c_2}{3y}$$

$$-(\frac{c_2}{3y})^3+y^3=-d_2$$

Then solve the quadratic equation with after $y^3=m$

$$-\frac{c^3_2}{27m}+m=-d_2$$

$$m^2+d_2.m-\frac{c^3_2}{27}=0$$

4. The qadratic equation can be solved via $m=s-\frac{d_2}{2}$

$$(s-\frac{d_2}{2})^2+d_2.(s-\frac{d_2}{2})-\frac{c^3_2}{27}=0$$

$$s^2=\frac{c^3_2}{27}+\frac{d^2_2}{2}=\Delta$$

Write $$s=\sqrt{\Delta}$$

$$m=\sqrt{\Delta}-\frac{d_2}{2}$$ $$y^3=m$$ $$y=\sqrt[3]{\sqrt{\Delta}-\frac{d_2}{2}}$$

$$p=-\frac{c_2}{3y}$$ $$p=-\frac{c_2}{3\sqrt[3]{\sqrt{\Delta}-\frac{d_2}{2}}}$$ $$p+y=z$$ $$x=\sqrt[3]{\sqrt{\Delta}-\frac{d_2}{2}}-\frac{c_2}{3\sqrt[3]{\sqrt{\Delta}-\frac{d_2}{2}}}-\frac{b_1}{3}$$