Solve quartic equation $a^4-6a^2b-8ac-3b^2=0$ Please help me to find roots of this quartic equation for a:
$$a^4-6a^2b-8ac-3b^2=0$$
Wolfram Alpha gave this result.
But may be there is simple way to get all a?
 A: Rewrite the equation by completing the squares
\begin{align}
a^4-6a^2b-8ac-3b^2
=(a^2+s)^2 -( 2s+6b)\left( a +\frac{4c}{2s+6b} \right)^2\tag1\\
\end{align}
where $s$ happens to satisfy $(s+b)^3=8(c^2-b^3)$, or
$s=2\sqrt[3]{c^2-b^3}-b$.
For notational convenience, denote
$$p=\frac12\sqrt{2s+6b}= \sqrt{\sqrt[3]{c^2-b^3}+b }\tag2$$
and then $s=2p^2-3b$.
Substitute them into (1) and then factorize to get
$$\left(a^2+2pa+2p^2-3b-\frac{2c}p\right) \left(a^2-2pa+2p^2-3b+\frac{2c}p\right)=0
$$
which yields the solutions
$$a=p\pm \sqrt{ 3b-p^2+\frac{2c}p },\>\>\> -p\pm \sqrt{ 3b-p^2-\frac{2c}p } $$
with $p$ given by (2).
A: $$\text{Given}\quad a^4-6a^2b-8ac-3b^2=0\quad\implies 3b^2+6a^2b -(a^4-8ac)=0$$
The solution for  $b$ using the quadratic equation is
$$b = \pm\frac{2 \sqrt{a^4 - 2 a c}}{\sqrt{3}} - a^2\quad\text{for}\quad a\in\mathbb{R}\land c\in\mathbb{R}\land b\in\mathbb{R}\iff a^4-2ac\ge0$$
Technically, this defines all values of $a$ but perhaps you seek integers.
Regular algebras solves for $c$ as $$c = \frac{a^4 - 6 a^2 b - 3 b^2}{8 a}$$
The second equation looks easier to work with (though you can try the first if you want) and experimentation in a spreadsheet offers these sample integer triples.
\begin{equation}
(1,-11,-37)\quad
(1,-7,-13)\quad
(1,-3,-1)\quad
(1,1,-1)\quad
(1,5,-13)\quad
(1,9,-37)\quad\\
(2,-8,1)\quad
(2,-4,4)\quad
(2,0,1)\quad
(2,4,-8)\quad
(2,8,-23)\quad
(2,12,-44)\quad\\
(3,-11,13)\quad
(3,-7,13)\quad
(3,-3,9)\quad
(3,1,1)\quad
(3,5,-11)\quad
(3,9,-27)\quad
(3,13,-47)\quad\\
(4,-8,26)\quad
(4,0,8)\quad
(4,8,-22)\quad\\
(5,5,-5)\quad\\
(6,-8,59)\quad
(6,-4,44)\quad
(6,0,27)\quad
(6,4,8)\quad
(6,8,-13)\quad
(6,12,-36)\quad\\
(7,-7,77)\quad\\
(8,-8,109)\quad
(8,0,64)\quad
(8,8,13)\quad
\end{equation}
WolframAlpha offer only the roots $a=0,b=0$ or
$a\ne0\land c = \frac{a^4 - 6 a^2 b - 3 b^2}{8 a}$ as shown here. This solution for $c$ (as I pointed out above) appears to have infinite integer solutions and, when $a=1$, it appears that these have a symmetry for values of $b$ around zero. Let me know if you have other questions about this.
