How to show $y^2,z^2,u^2$ are roots of $(1)$ Euler's solution for the general quartic are as follows:

From the depressed quartic, $x^4+px^2+qx+r=0$, assume $x=y+z+u$ and it may be shown that $u^2,y^2,z^2$ are roots of the cubic$$t^3+\dfrac q2t^2+\dfrac {q^2-4s}{16}t-\dfrac {r^2}{64}=0\tag1$$
$$\vdots$$

However, how would you go about showing that $u^2,z^2,y^2$ are roots of $(1)$? I originally started by writing $(1)$ as$$(t-u^2)(t-y^2)(t-z^2)\tag2$$
And expanding to get $t^3-(u^2+y^2+z^2)t^2+(u^2y^2+u^2z^2+y^2z^2)t-u^2y^2z^2$. However, now I don't know what to do. Any pointers on what to do now?

That's Decartes–Euler solution
\begin{align*}
  x &= u+v+w \\[5pt]
  x^2 &= u^2+v^2+w^2 + 2(uv+vw+wu) \\[5pt]
  x^4 &= (u^2 + v^2 + w^2)^2 + 4(u^2+v^2+w^2)(uv+vw+wu) \\
  &\quad +4(u^2 v^2+v^2 w^2+w^2 u^2)+8uvw(u+v+w) \\[5pt]
  0 &= x^4+px^2+qx+r \\[5pt]
  0 &= (u^2+v^2+w^2)^2+(uv+vw+wu)[\color{blue}{4(u^2+v^2+w^2)+2p}] \\
  & \quad +(u+v+w)(\color{red}{8uvw+q})+4(u^2 v^2+v^2 w^2+w^2 u^2) \\
  & \quad +p(u^2+v^2+w^2)+r \\
\end{align*}
You have two degree of freedom on choosing the combination of $u$, $v$ and $w$.
Eliminate the terms with factor of $\, u+v+w \,$ or $\, uv+vw+wu \,$ by taking
$$
\left \{
  \begin{array}{rcL}
    \color{blue}{4(u^2+v^2+w^2)+2p} &=& 0 \\[8pt]
    \color{red}{8uvw+q} &=& 0
  \end{array}
\right. \quad \implies \quad
\left \{
  \begin{array}{rcL}
    u^2+v^2+w^2 &=& -\dfrac{p}{2} \\[5pt]
    u^2v^2w^2 &=& \dfrac{q^2}{64}
  \end{array}
\right.$$
Also from the original quartic,
$$(u^2+v^2+w^2)^2+4(u^2 v^2+v^2 w^2+w^2 u^2)+p(u^2+v^2+w^2)+r=0$$
$$u^2 v^2+v^2 w^2+w^2 u^2=\frac{p^2}{16}-\frac{r}{4}$$
Now we can form an equation with roots $u^2$, $v^2$ and $w^2$, namely
$$t^3+\frac{p}{2}t^2+\frac{p^2-4r}{16}t-\frac{q^2}{64}=0$$
