I tried deriving the solutions of a cubic polynomial using a method with which I derived solutions for a quadratic polynomial.
Here is the method for quadratics:
Let $P(x) = ax^2+bx+c$. The value of $x$ for which $\frac{dP(x)}{dx} = 0$ is $-\frac{b}{2a}$. We will call it $\alpha$. It can easily be shown that $P(\alpha+x) = P(\alpha-x)$. Geometrically it means that $P(x)$ has the line of symmetry $ y=\alpha$. Meaning that the polynomial $P(x+\alpha)$ has the line of symmetry $y=0$ and therefore its solutions are opposite of each other. Which means that $P(x+\alpha) = (x+x_1)(x-x_1) = x^2 - x_1^2$ where $x_1$ and $-x_1$ are its solutions, which can be calculated more easily. Knowing that I made the substitution $x=y+\alpha$ and it resulted in: $$ P(x) = ay^2+c+\alpha\frac{b}{2} = ay^2+\beta $$
So it follows $$P(x) = 0 \iff ay^2+\beta = 0 \iff y = \pm\sqrt{-\frac{\beta}{a}} \iff $$
$$ x = \pm\sqrt{-\frac{\beta}{a}} + \alpha = \frac{-b \pm \sqrt{b^2-4ac}}{2a} $$
My attempt for cubics:
$P(x) = ax^3+bx^2+cx+d$. The value of $x$ for which $\frac{d^2P(x)}{dx^2} = 0$ is $-\frac{b}{3a}$ which I call $\alpha$. There is the symmetry $P(\alpha+x) = -P(\alpha-x)$So it seems natural to make the substitution $x=y+\alpha$. This leads to :
$$ P(x) = ay^3+y(c+b\alpha )+c\alpha+d $$
And here I'm stuck.