Solve {l}{2a+b/c=2b+c/a=2c+a/b=k}{a+b+cneq0}

$$\left. \begin{array} { l } { \frac { 2 a + b } { c } = \frac { 2 b + c } { a } = \frac { 2 c + a } { b } = k } \\ { a + b + c \neq 0 } \end{array} \right.$$

Solve for a, b, c (complex solution)

$$\left\{\begin{matrix}\\a=\frac{-\sqrt{b^{2}+16bc+8c^{2}}-b}{4}\text{, }b\neq -\frac{c}{2}\text{, }c\neq 0\text{; }a=\frac{\sqrt{b^{2}+16bc+8c^{2}}-b}{4}\text{, }b\neq -\frac{c}{2}\text{, }c\neq 0\text{, }&\text{unconditionally}\\a=\frac{-\sqrt{b^{2}+16bc+8c^{2}}-b}{4}\text{, }b\in \mathrm{C}\text{, }c\in \mathrm{C}\text{, }&c\neq 0\text{ and }arg(c)\geq \pi \\a=\frac{\sqrt{b^{2}+16bc+8c^{2}}-b}{4}\text{, }b\in \mathrm{C}\text{, }c\in \mathrm{C}\text{, }&c\neq 0\text{ and }|\frac{arg(c^{2})}{2}-arg(-c)|\geq \pi \end{matrix}\right.$$

Solve for a, b, c

$$\left\{\begin{matrix}\\a=\frac{-\sqrt{b^{2}+16bc+8c^{2}}-b}{4}\text{, }b\geq -2\sqrt{14}c-8c\text{, }c<0\text{; }a=\frac{-\sqrt{b^{2}+16bc+8c^{2}}-b}{4}\text{, }b\leq -2\sqrt{14}|c|-8c\text{, }c<0\text{; }a=\frac{-\sqrt{b^{2}+16bc+8c^{2}}-b}{4}\text{, }b\leq 2\sqrt{14}c-8c\text{, }c<0\text{; }a=\frac{\sqrt{b^{2}+16bc+8c^{2}}-b}{4}\text{, }b\leq -2\sqrt{14}c-8c\text{, }c>0\text{; }a=\frac{\sqrt{b^{2}+16bc+8c^{2}}-b}{4}\text{, }b\geq 2\sqrt{14}|c|-8c\text{, }c>0\text{; }a=\frac{\sqrt{b^{2}+16bc+8c^{2}}-b}{4}\text{, }b\in [2\sqrt{14}c-8c,0)\text{, }c>0\text{, }&\text{unconditionally}\\a=\frac{-\sqrt{b^{2}+16bc+8c^{2}}-b}{4}\text{, }b\in \mathrm{R}\text{, }c<0\text{; }a=\frac{-\sqrt{b^{2}+16bc+8c^{2}}-b}{4}\text{, }b\in \mathrm{R}\setminus (-\frac{c}{2}\cup (-\infty,2\sqrt{14}|c|-8c)\cup (-2\sqrt{14}|c|-8c,2\sqrt{14}|c|-8c)\text{, }c\neq 0)\text{, }c>0\text{; }a=\frac{\sqrt{b^{2}+16bc+8c^{2}}-b}{4}\text{, }b\in \mathrm{R}\setminus ((-\infty,2\sqrt{14}c-8c)\cup [0,\infty)\cup (-\infty,-\frac{c}{2}])\text{, }c>0\text{, }&c\geq 0\\a=\frac{\sqrt{b^{2}+16bc+8c^{2}}-b}{4}\text{, }b\in \mathrm{R}\text{, }c>0\text{; }a=\frac{-\sqrt{b^{2}+16bc+8c^{2}}-b}{4}\text{, }b\in \mathrm{R}\setminus (-\frac{c}{2}\cup (2\sqrt{14}c-8c,\infty)\cup (-2\sqrt{14}|c|-8c,\infty)\cup (-2\sqrt{14}|c|-8c,-\frac{c}{2}]\text{, }c>0\cup (-\infty,-\frac{c}{2}]\text{, }c>0)\text{, }c<0\text{; }a=\frac{\sqrt{b^{2}+16bc+8c^{2}}-b}{4}\text{, }b\in \mathrm{R}\setminus (-\frac{c}{2}\cup (-2\sqrt{14}|c|-8c,\infty)\cup (-2\sqrt{14}|c|-8c,2\sqrt{14}|c|-8c)\text{, }c>0)\text{, }c<0\text{, }&c\leq 0\\a=\frac{-\sqrt{b^{2}+16bc+8c^{2}}-b}{4}\text{, }b\in \mathrm{R}\setminus ((-\infty,-2\sqrt{14}c-8c)\text{, }c>0\cup (-2\sqrt{14}|c|-8c,\infty)\text{, }c\geq 0\cup (-\infty,-\frac{c}{2}]\cup (2\sqrt{14}c-8c,\infty)\text{, }c>0\cup (-\infty,0]\text{, }c<0\cup (-2\sqrt{14}|c|-8c,0]\text{, }c>0\cup (-\infty,2\sqrt{14}|c|-8c))\text{, }c<0\text{; }a=\frac{\sqrt{b^{2}+16bc+8c^{2}}-b}{4}\text{, }b\in \mathrm{R}\setminus ((-\infty,-\frac{c}{2}]\cup (-2\sqrt{14}|c|-8c,\infty)\text{, }c\geq 0\cup (-\infty,2\sqrt{14}|c|-8c))\text{, }c<0\text{, }&c<0\\a=\frac{-\sqrt{b^{2}+16bc+8c^{2}}-b}{4}\text{, }b\in \mathrm{R}\setminus ((-\infty,2\sqrt{14}|c|-8c)\text{, }c\leq 0\cup [-\frac{c}{2},\infty)\cup (-2\sqrt{14}|c|-8c,\infty))\text{, }c>0\text{; }a=\frac{\sqrt{b^{2}+16bc+8c^{2}}-b}{4}\text{, }b\in \mathrm{R}\setminus ([-\frac{c}{2},\infty)\cup (-2\sqrt{14}|c|-8c,2\sqrt{14}|c|-8c)\text{, }c\neq 0\cup [-\frac{c}{2},2\sqrt{14}|c|-8c)\text{, }c<0\cup [0,2\sqrt{14}c-8c)\text{, }c<0\cup [-\frac{c}{2},2\sqrt{14}c-8c)\text{, }c<0\cup [0,\infty)\cup (-2\sqrt{14}|c|-8c,\infty))\text{, }c>0\text{, }&c>0\end{matrix}\right.$$