Is there anything like “cubic formula”?

Question

Just like if we have any quadratic equation which has complex roots, then we are not able to factorize it easily. So we apply quadratic formula and get the roots.

Similarly if we have a cubic equation which has two complex roots (which we know conjugate of each other) and one fractional root, then we are not able to find its first root by hit & trial.

So my question is like quadratic formula, is there exist any thing like cubic formula which help in solving cubic equations?

For example, I have an equation $$2x^3+9x^2+9x-7=0\tag{1}$$ and I have to find its solution which I am not able to find because it has no integral solution. Its solutions are $\dfrac {1}{2}$, $\dfrac{-5\pm \sqrt{3}i}{2} $, I know these solutions because this equation is generated by myself.

So how can I solve equations like these?

Also while typing this question, I thought about the derivation of quadratic formula, which is derived by completing the square method.

So I tried to apply ‘completing the cube’ method on the general equation $ax^3+bx^2+cx+d=0$ but it didn't help.

So please help me in finding a cubic formula or to solve the equations like given in example by an alternative method.

Answer 1

Yes, we do have a cubic formula! By Cardan's Method...

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Cardan's Method: To solve the general cubic$$x^3+ax^2+bx+c=0\tag{i}$$ Remove the $ax^2$ term by substituting $x=\dfrac {y-a}3$. Let the transformed equation be$$y^3+qy+r=0\tag{ii}$$ To solve this depressed cubic, substitute $y=u+v$ to get$$u^3+v^3+(3uv+q)(u+v)+r=0\tag{iii}$$ Put $3uv+q=0$ to get $u=-\dfrac q{3v}$ and substituting this back gives a quadratic in $v^3$. The roots of the quadratic are equal to $u^3,v^3$ respectively. And from our substitution, we get a root as$$y=\left\{-\frac r2+\sqrt{\frac {r^2}4+\frac {q^3}{27}}\right\}^{\frac 13}+\left\{-\frac r2-\sqrt{\frac {r^2}4+\frac {q^3}{27}}\right\}^{\frac 13}\tag{iv}$$ With the other two roots found with the cube roots of unity.

To find the original root of $(\text i)$, substitute $y$ into your transformation.

Answer 2

Absolutely!

Suppose

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

, $i^2=-1$, and exactly one of the following is true.

$w_1=w_2=1$

or $w_1=\frac{-1+i\sqrt3}{2}$,and$w_2=\frac{-1-i\sqrt3}{2}$

or $w_1=\frac{-1-i\sqrt3}{2}$, and $w_2=\frac{-1+i\sqrt3}{2}$

then

$$x=-\frac{b}{3a}+w_1\sqrt[3]{\frac{9abc-2b^3-27a^2d}{54a^3}+\sqrt{(\frac{9abc-2b^3-27a^2d}{54a^3})^2+(\frac{3ac-b^2}{9a^2})^3}}+w_2\sqrt[3]{\frac{9abc-2b^3-27a^2d}{54a^3}-\sqrt{(\frac{9abc-2b^3-27a^2d}{54a^3})^2+(\frac{3ac-b^2}{9a^2})^3}}$$

Credit goes to Girolamo Cardano.

This formula can be easily memorized once you notice certain patterns.

Answer 3

One method is to depress the cubic, then apply trig functions.

$$0=sx^3+tx^2+ux+v$$

Divide both sides by $s$ to get:

$$0=x^3+ax^2+bx+c$$

Let $x=y-\frac a3$ to get

$$0=y^3+\underbrace{\left(b-\frac{a^2}3\right)}_dy+\underbrace{c-\frac{ab}3+\frac{2a^3}{27}}_e=y^3+dy+e$$

If $d>0$, then use the trigonometric identity:

$$\sinh(3\theta)=4\sinh^3(\theta)+3\sinh(\theta)$$

where

$$\sinh(\theta)=\frac{e^\theta-e^{-\theta}}2$$

We exploit this identity by letting $y=fz$ and multiplying both sides by $g$ to get

$$0=f^3gz^3+dfgz+eg$$

$$\begin{cases}4=f^3g\\3=dfg\end{cases}\implies\begin{cases}f=2\sqrt{\frac d3}\\g=\frac{3\sqrt3}{2d^{3/2}}\end{cases}$$

$$0=4z^3+3z+\frac{3e\sqrt3}{2d^{3/2}}=\sinh(3\operatorname{arcsinh}(z))+\frac{3e\sqrt3}{2d^{3/2}}$$

$$\implies\sinh(3\operatorname{arcsinh}(z))=-\frac{3e\sqrt3}{2d^{3/2}}$$

$$\implies z=-\sinh\left(\frac13\operatorname{arcsinh}\left(\frac{3e\sqrt3}{2d^{3/2}}\right)\right)$$

$$\implies x=-2\sqrt{\frac d3}\sinh\left(\frac13\operatorname{arcsinh}\left(\frac{3e\sqrt3}{2d^{3/2}}\right)\right)-\frac a3$$

If $d<0$, use $\cos(3\theta)$ or $\cosh(3\theta)$ and respective triple angle formulas.