derivation of the cubic formula In my research I came across the following derivation of the cubic formula that was almost complete. However at the end it says to make  the substitution $u=z^3$ which would determine the roots. I am not quite sure how to do this and then how to determine the roots and I am looking for some help.
A general form for the cubic equation is,
$$ax^3+bx^2+cx+d=0 \tag{1}$$
To find the roots of this equation we first try to get rid of the quadratic term $x^2$. The substitution $x=y-\dfrac{b}{3a}$ helps in achieving our goal. This results in, $$ay^3+\left(c-\dfrac{b^2}{3a}\right)y+\left(d+\dfrac{2b^3}{27a^2}-\dfrac{bc}{3a}\right)=0\tag{2}$$ which we transform into the following, $$y^3+\dfrac{1}{a}\left(c-\dfrac{b^2}{3a}\right)y+\dfrac{1}{a}\left(d+\dfrac{2b^3}{27a^2}-\dfrac{bc}{3a}\right)=0\tag{3}$$ Upon assuming $e=\dfrac{1}{a}\left(c-\dfrac{b^2}{3a}\right)$ and $f=\dfrac{1}{a}\left(d+\dfrac{2b^3}{27a^2}-\dfrac{bc}{3a}\right)$ we get the equation as, $$y^3+ey+f=0\tag{4}$$ We reduce this equation by the substitution $y=z+\dfrac{s}{z}$ and choosing $s=-\dfrac{e}{3}$ we obtain the simplified equation as, $$z^6+fz^3-\dfrac{e^3}{27}=0\tag{5}$$ What only remains is to make the substitution $u=z^3$.
 A: The cubic formula was derived from a series of substitutions. It's probably better to memorize the process of deriving the formula, rather than memorizing the actual formula.
Starting with the general cubic, make a substitution such that the squared term is removed. Let's denote the depressed cubic as$$x^3+qx+r=0$$Now, set $x=y+z$. Expanding and factoring gives$$y^3+z^3+(3yz+q)(y+z)+r=0$$Setting $3yz+q=0$ gives $$y=-\frac q{3z}$$If we substitute that in, we get a quadratic in $y^3$. Using the quadratic formula, we get the roots of $y^3$ and $z^3$ respectively. Therefore, we have the solution as$$x_1=\color{blue}{\left\{-\frac r2+\sqrt{\frac {r^2}4+\frac {q^3}{27}}\right\}^{\tfrac 13}+\left\{-\frac r2-\sqrt{\frac {r^2}4-\frac {q^3}{27}}\right\}^{\tfrac 13}}$$

The other solutions can be found using the cube roots of unity.$$\omega=-\frac 12+\frac {i\sqrt{3}}2\qquad\omega^2=-\frac 12-\frac {i\sqrt3}2$$So$$\begin{align*} & x_2=\color{blue}{\omega\left\{-\frac r2+\sqrt{\frac {r^2}4+\frac {q^3}{27}}\right\}^{\frac 13}+\omega^2\left\{-\frac r2-\sqrt{\frac {r^2}4+\frac {q^3}{27}}\right\}^{\frac 13}}\\ & x_3=\color{blue}{\omega^2\left\{-\frac r2+\sqrt{\frac {r^2}4+\frac {q^3}{27}}\right\}^{\frac 13}+\omega\left\{-\frac r2-\sqrt{\frac {r^2}4+\frac {q^3}{27}}\right\}^{\frac 13}}\end{align*}$$
