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?
 A: 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
$$t^3+\frac{p}{2}t^2+\frac{p^2-4r}{16}t-\frac{q^2}{64}=0$$
A: COMMENT.-Euler himself used the following equation to apply his method:
 $$x^4-8x^3+14x^2+4x-8=0\tag1$$
I reproduce this solution because it seems to me illustrative for who likes the subject and of certain historical interest. 
First a synthesis of the Euler method to solve the quartic.
$$***$$


*

*$Ax^4+Bx^3+Cx^2+Dx+E=0$ is reduced to $x^4-ax^2-bx-c=0$ through the change of variables $y=x-\dfrac{B}{4A}$ similar to the known change for the cubic equation.

*Euler assumes that a quartic of the form $x^4-ax^2-bx-c=0$ has a solution of the form $$x=\sqrt p+\sqrt q+\sqrt r\tag{2}$$

*Taking twice the square in $(2)$ he got
$$x^4-2fx^2-8\sqrt h x-(4g-f^2)=0$$ where $f=p+q+r$, $g=pq+pr+qr$ and $h=pqr$. (It is clear that $p,q,r$ are roots of $t^3-fx^2+gx-h=0$)

*Comparing (3) with (2) he got $f=\dfrac a2$, $h=\dfrac{b^2}{64}$ and $g=\dfrac{4c+a^2}{16}$
$$***$$


$$\color{blue}{\text{ Solve the equation } x^4-8x^3+14x^2+4x-8=0}$$
The reduced equation through $y=x+2$ is $$x^4+10x^2-4x+8=0$$
Since $f=\dfrac{10}{2}=5,h=\dfrac 14$ and $g=\dfrac{17}{4}$, $p,q,r$ are roots of the equation $$t^3-5t^2+\dfrac{17}{4}t-\dfrac 14=0\tag3$$ whose solutions are
$$p=1,\space q=\frac{4+\sqrt{15}}{2},\space  r= \frac{4-\sqrt{15}}{2}\tag4$$
Hence
$$\sqrt p=\pm1,\space \sqrt q=\sqrt{\frac{4+\sqrt{15}}{2}},\space \sqrt r=\sqrt{\frac{4-\sqrt{15}}{2}}\tag5$$ Euler noticed that $(\sqrt5\pm\sqrt3)^2=\sqrt{8\pm 2\sqrt{15}}$ so he simplified
$$\sqrt p=\pm1,\space \sqrt q=\pm\frac{\sqrt5+\sqrt3}{2},\space \sqrt r=\pm\frac{\sqrt5-\sqrt3}{2}\tag6$$ Furthermore, because of $\dfrac q8=\sqrt h=\sqrt p\sqrt q\sqrt r\gt 0$,  Euler had to give signs to the three square roots so that his product be positive which gives
$$\begin{align*} & \sqrt p+\sqrt q+\sqrt r=1+\sqrt5\\ & \sqrt p-\sqrt q-\sqrt r=1-\sqrt5\\ & -\sqrt p+\sqrt q-\sqrt r=-1+\sqrt5\\ & -\sqrt p-\sqrt q+\sqrt r=-1-\sqrt5\end{align*}\tag7$$ Thus, taking into account $y=x+2$ Euler got the four roots of his equation $$x_{1,2}=3\pm\sqrt5\\x_{3,4}=1\pm\sqrt3$$
