<s>[INST] If $\alpha$ be a multiple root of the order 3 of the equation $x⁴+bx²+cx+d=0$ then $α= 8d/3c​$ I found this statement online here and I think this statement is false.I would first  prove a general statement and then I would use it to find $\alpha$ in order to discard the statement given in online link.
Theorem:If the equation $x^4+ax^3+bx^2+cx+d=0$ has three equal roots, show that each of them is equal to $\frac{6c-ab}{3a^2-8b}$​
Proof;Three roots of the given equation are same hence, the roots can be assumed as $\alpha,\alpha,\alpha,\beta$
Here, $S_1=3α+β=−a;S_2
=3α(α+β)=b;S_3=α^2(α+3β)=−c;S_4=α^3β=d$
We need to evaluate the value of $\frac{6c-ab}{3a^2-8b}$​.
$6c−ab=α(3α^2−6αβ+3β^2)$
$3a^2−8b=3α^2−6αβ+3β^2$
Therefore,$\frac{6c-ab}{3a^2-8b}=\frac{α(3α^2−6αβ+3β^2)}{(3α^2−6αβ+3β^2)}=\alpha$
Now according to this ,putting $a=0$  we get $\alpha=\frac{-3c}{4b}$ which is not same as $\frac{8d}{3c}$ as claimed in online site.
My question is: Is the claim in online site wrong?or I have gone wrong somewhere in my proof.
Any help is much appreciated!
 [/INST] Verification of the Theorem
Expanding
$$
(x-u)^3(x-v)=x^4\ \overbrace{-(3u+v)}^{\large a}x^3\ \overbrace{+3u(u+v)}^{\large b}x^2\ \overbrace{-u^2(u+3v)}^{\large c}x\overbrace{\ +u^3v\ \ }^{\large d}\tag1
$$
The first two derivatives are $0$ at $u$
$$
u^4+au^3+bu^2+cu+d=0\tag2
$$
$$
4u^3+3au^2+2bu+c=0\tag3
$$
$$
12u^2+6au+2b=0\tag4
$$
Substituting $v=d/u^3$ into $c+u^2(u+3v)=0$ gives
$$
u^4+cu+3d=0\tag5
$$
Writing $(2)$-$(5)$ as a matrix equation yields
$$
\begin{align}
0
&=\begin{bmatrix}
1&-\frac a4&\frac{3a^2-4b}{48}&-1
\end{bmatrix}
\begin{bmatrix}
1&a&b&c&d\\
0&4&3a&2b&c\\
0&0&12&6a&2b\\
1&0&0&c&3d
\end{bmatrix}
\begin{bmatrix}
u^4\\u^3\\u^2\\u\\1
\end{bmatrix}\tag{6a}
\\
&=\begin{bmatrix}
0&0&0&\frac{3a^3-8ab}8&\frac{3a^2b-4b^2-6ac-48d}{24}
\end{bmatrix}
\begin{bmatrix}
u^4\\u^3\\u^2\\u\\1
\end{bmatrix}\tag{6b}
\end{align}
$$
Since $b^2+12d=3ac$, $\text{(6b)}$ gives
$$
\begin{align}
u
&=\frac{3a^2b-4b^2-6ac-48d}{24ab-9a^3}\tag{7a}\\
&=\frac{3a^2b-18ac}{24ab-9a^3}\tag{7b}\\
&=\frac{ab-6c}{8b-3a^2}\tag{7c}
\end{align}
$$

When $\boldsymbol{a=0}$
Plugging $a=0$ into $\text{(7c)}$ gives
$$
u=-\frac{3c}{4b}\tag8
$$
If we use $a=0$, that is, $v=-3u$, we get $b=-6u^2$, $c=8u^3$, and $d=-3u^4$, we get not only that $u=-\frac{3c}{4b}$, as given in $(8)$, but also that
$$
u=-\frac{8d}{3c}\tag9
$$
which says that the problem in question is a factor of $-1$ off.
</s>