Question: I was preparing for an entrance exam and found a question a sample paper that I could not solve.

Given, $$ a + b + c = 0 $$
Find the value of $$ a^2 (b + c) + b^2 (c + a) + c^2 (a + b) \over abc $$
I tried to do this many different ways. Some of my findings are -
$$ a^2 (b + c) + b^2 (c + a) + c^2 (a + b) \over abc $$
$$= {ab + ac \over bc} + {bc+ab \over ac} + {ac+bc \over ab} $$
$$ = {(ab)^2 +(ac)^2 +2a^2 bc+(ab)^2 +(bc)^2 +2ab^2 c+(bc)^2 +(ac)^2+2abc^2\over (ab)(ac)(bc)}$$
$$ = {(ab +ac)^2 +(ab+bc)^2 +(bc+ac)^2\over (ab)(ac)(bc)}$$
Am I headed the right way?

Answer: 
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.
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.