Solve 3b^2c-12bc^2=4b^2c^2(2b-3c)

$$3 b ^ { 2 } c - 12 b c ^ { 2 } = 4 b ^ { 2 } c ^ { 2 } ( 2 b - 3 c )$$

Solve for b (complex solution)

$$\left\{\begin{matrix}\\b=0\text{, }&\text{unconditionally}\\b=-\frac{\sqrt{3\left(48c^{4}-104c^{2}+3\right)}-12c^{2}-3}{16c}\text{; }b=\frac{\sqrt{3\left(48c^{4}-104c^{2}+3\right)}+12c^{2}+3}{16c}\text{, }&c\neq 0\\b\in \mathrm{C}\text{, }&c=0\end{matrix}\right.$$

Solve for c (complex solution)

$$\left\{\begin{matrix}\\c=0\text{, }&\text{unconditionally}\\c=\frac{\sqrt{4b^{4}+3b^{2}+9}+2b^{2}+3}{6b}\text{; }c=\frac{-\sqrt{4b^{4}+3b^{2}+9}+2b^{2}+3}{6b}\text{, }&b\neq 0\\c\in \mathrm{C}\text{, }&b=0\end{matrix}\right.$$

Solve for b

$$\left\{\begin{matrix}\\b=0\text{, }&\text{unconditionally}\\b=-\frac{\sqrt{3\left(48c^{4}-104c^{2}+3\right)}-12c^{2}-3}{16c}\text{; }b=\frac{\sqrt{3\left(48c^{4}-104c^{2}+3\right)}+12c^{2}+3}{16c}\text{, }&\left(c\neq 0\text{ and }|c|\leq \frac{\sqrt{6}}{3}-\frac{\sqrt{15}}{6}\right)\text{ or }|c|\geq \frac{\sqrt{6}}{3}+\frac{\sqrt{15}}{6}\\b\in \mathrm{R}\text{, }&c=0\end{matrix}\right.$$

Solve for c

$$\left\{\begin{matrix}\\c=0\text{, }&\text{unconditionally}\\c=\frac{\sqrt{4b^{4}+3b^{2}+9}+2b^{2}+3}{6b}\text{; }c=\frac{-\sqrt{4b^{4}+3b^{2}+9}+2b^{2}+3}{6b}\text{, }&b\neq 0\\c\in \mathrm{R}\text{, }&b=0\end{matrix}\right.$$