General Solution to Cubic Equation

In a recent post titled All Cubic Polynomials Are Point Symmetric, the key point was that the $x$ coordinate of the point of symmetry is given by: $$x=\frac{-b}{3a}$$ and this fact also plays a crucial role in devising a solution to the general cubic equation $ax^3+bx^2+cx+d=0$. It enables the point of symmetry in the graph of the polynomial to be translated horizontally so it lies on the $y$ axis and the new equation is much easier to solve because the $x^2$ term is removed.

Knowledge of the quadratic formula is older than the Pythagorean Theorem. Solving a cubic equation, on the other hand, was the first major success story of Renaissance mathematics in Italy. The solution was first published by Girolamo Cardano (1501-1576) in his Algebra book Ars Magna. Source

The technique finds a single real root and the other two roots (real or complex) can be found by polynomial division and the use of the quadratic formula. The first step in the solution is to horizontally translate the point of symmetry so it lies on the $y$ axis. This is achieved by means of the substitution: $$x=y-\frac{b}{3a}$$ This produces: $$a(y-\frac{b}{3a})^3+b(y-\frac{b}{3a})^2+c(y-\frac{b}{3a})+d=0$$ Multiplying this out gives $$ay^3+(c-\frac{b^2}{3a})y+(d+\frac{2b^2}{27a^2}-\frac{bc}{3a})=0$$ After division by $a$, this equation is of the general form $y^3+Ay=B$ and was solved by Scipione del Ferro (1465-1526).

Consider a specific cubic equation e.g. $x^3-15x^2+81x-175=0$.

Here the substitution will be $x=y+5$ because $a=1$ and $b=-15$ leading to $y^3+6y-20=0$. The situation is shown in the graph below, where it can be seen that the root of the original equation is 7 and that of the translated equation is 2 (although algebraically we don't know that yet):

We need to find $s$ and $t$ such that $3st=A$ and $s^3-t^3=B$.

It can be shown that $y=s-t$ is then a solution of $y^3+Ay=B$: $$(s-t)^3+3st(s-t)=s^3-t^3$$ $$(s^3-3s^2t+3st^2-t^3)+(3s^2t-3st^2)=s^3-t^3$$ Let's solve $3st=A$ and $s^3-t^3=B$ for the simplified cubic $y^3+6y=20$.

$3st=6$ and so $s=2/t$.

Substituting into $s^3-t^3=20$ gives $(2/t)^3-t^3=20$.

Multiplying by $t^3$ gives $t^6+20t^3-8=0$

Taking the positive root $t^3=-10+\sqrt{108}$ and $t=\sqrt[3]{-10+\sqrt{108}}$

Now $s^3=20+t^3$ and so $s^3=20+(-10+\sqrt{108})=10+\sqrt{108}$

This means that $s=\sqrt[3]{10+\sqrt{108}}$ and then, because $y=s-t$, we have:

$y=\sqrt[3]{10+\sqrt{108}}-\sqrt[3]{-10+\sqrt{108}}$

But $x=y+5$ and so $x=\sqrt[3]{10+\sqrt{108}}-\sqrt[3]{-10+\sqrt{108}}+5$

Remarkably, this formidable root (the only real root) evaluates to 7!

So that means $\sqrt[3]{10+\sqrt{108}}-\sqrt[3]{-10+\sqrt{108}}=2$. Why is it so?

Well, here is how it all comes about:

$\sqrt[3]{10 + \sqrt{108}} - \sqrt[3]{\sqrt{108} - 10}$

$=\sqrt[3]{10 + 6 \sqrt{3}} - \sqrt[3]{6 \sqrt{3} - 10}$

$=\sqrt[3]{1^3+3\sqrt{3}+3(\sqrt{3})^2+(\sqrt{3})^3}-\sqrt[3]{-1^3+3\sqrt{3}-3(\sqrt{3})^2+(\sqrt{3})^3}$

$=\sqrt[3]{(1+\sqrt{3})^3}-\sqrt[3]{(-1+\sqrt{3})^3}$

$=1+\sqrt{3}+1-\sqrt{3}$

$=2$

Reference: The "Cubic Formula"