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!
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 $y= z - \frac p{3z}$.
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
$$z^6 + qz^3 - \left(\dfrac{p}{3}\right)^3 = 0$$
