$$\frac{ x+a }{ 3 } = \frac{ { \left(x-b \right) }^{ 2 } }{ 3x-a } + \frac{ 3ab-3a { b }^{ 2 } }{ 9x-3a }$$
$\left\{\begin{matrix}b=\frac{\sqrt{3\left(2x-a\right)\left(6x-4a^{2}+a\right)}+3a-6x}{6\left(a-1\right)}\text{; }b=\frac{-\sqrt{3\left(2x-a\right)\left(6x-4a^{2}+a\right)}+3a-6x}{6\left(a-1\right)}\text{, }&a\neq 1\text{ and }x\neq \frac{a}{3}\\b=-\frac{1}{3}\approx -0.333333333\text{, }&a=1\text{ and }x\neq \frac{1}{2}\text{ and }x\neq \frac{1}{3}\\b\in \mathrm{C}\text{, }&x=\frac{1}{2}\text{ and }a=1\end{matrix}\right.$
$\left\{\begin{matrix}b=\frac{\sqrt{3\left(2x-a\right)\left(6x-4a^{2}+a\right)}+3a-6x}{6\left(a-1\right)}\text{; }b=\frac{-\sqrt{3\left(2x-a\right)\left(6x-4a^{2}+a\right)}+3a-6x}{6\left(a-1\right)}\text{, }&a\neq 1\text{ and }\left(x\leq -\frac{\sqrt{a^{4}-2a^{3}+a^{2}}}{3}+\frac{a^{2}}{3}+\frac{a}{6}\text{ or }x\geq \frac{\sqrt{a^{4}-2a^{3}+a^{2}}}{3}+\frac{a^{2}}{3}+\frac{a}{6}\right)\text{ and }x\neq \frac{a}{3}\\b=-\frac{1}{3}\approx -0.333333333\text{, }&a=1\text{ and }x\neq \frac{1}{2}\text{ and }x\neq \frac{1}{3}\\b\in \mathrm{R}\text{, }&x=\frac{1}{2}\text{ and }a=1\end{matrix}\right.$