How to deduce the Vieta's substitution $y = z-\frac {p}{3z}$ for cubic equation $y^3+py +q=0$ In deriving the formula of cubic equations, Vieta substituted the following
$$y = z-\frac {p}{3z}$$
for the depressed cubic equation
$$y^3+py +q=0$$
and transformed it into a quadratic one.
My question: How did he get that substitution, or how did he know that by substituting $y = z-\frac {p}{3z}$, he could turn cubic into quadratic?
Please help me!
 A: The quadratic equation in $u$ has the equivalent form below
$$u+ \frac a{u}+ b=0$$
Determine below the substitution that transforms the depressed cubic equation $y^3+p y+q=0$ into a quadratic one, i.e.
\begin{align}
y^3+p y +q &= u+\frac a{u}+ b\\
& = \left(u^{\frac13}+\frac {a^{\frac13}}{u^{\frac13}}\right)\left( u^{\frac23 }+\frac {a^{\frac23}}{u^{\frac23}} -a^{\frac13}\right)+b\\
&= \left(u^{\frac13}+\frac {a^{\frac13}}{u^{\frac13}}\right)
\left( \left(u^{\frac13 }+\frac {a^{\frac13}}{u^{\frac13}}\right)^2 - 3a^{\frac13}\right)+b\\
&= \left(u^{\frac13}+\frac {a^{\frac13}}{u^{\frac13}}\right)^3
- 3a^{\frac13} \left(u^{\frac13 }+\frac {a^{\frac13}}{u^{\frac13}}\right)+b\\
\end{align}
Compare the two sides to get $b=q$, $a^{\frac13}=-\frac p3$  and $y = u^{\frac13}+\frac {a^{\frac13}}{u^{\frac13} }$. Then, let $u(z)= z^3$ to obtain the Vieta’s substitution $y= z - \frac p{3z}$.
A: This is a sligntly different way of looking at your problem and getting the same results.
You have an equation of the form $$t^3 + pt + q=0 \tag{A.}$$
Let $t=z-w$. Then
$\begin{align}
   t^3 &= z^3 - 3z^2V + 3zw^2 - w^3 \\
   t^3 &= 3zw(z - w) + (z^3 - w^3) \\
   t^3 &= 3zwt + (z^3 - w^3)
\end{align}$
Hence
$$t^3 - 3zwt - (z^3-w^3) = 0 \tag{B.}$$
Comparing (A.) and (B.), we get
\begin{align}
   p &= -3zw \\
   q &= -z^3 + w^3 \\
\end{align}
Rewriting the second equation as
$$z^3 + q - w^3 = 0$$
And substituting $w = -\frac{p}{3z}$, we get
$z^3 + q - \left(\dfrac{p}{3z}\right)^3 = 0$, which becomes
$$z^6 + qz^3 - \left(\dfrac{p}{3}\right)^3 = 0$$
