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Expanding the equation, we get $\dfrac{a^2+2a\sqrt{b}+b}{c^2} = a + \dfrac{\sqrt{b}}{c}$ . Separating the irrational part from the rational part, we get the system of two equations: $\begin{cases} \dfrac{a^2+b}{c^2} = a\\ \\ \dfrac{2a\sqrt{b}}{c^2} = \dfrac{\sqrt{b}}{c} \end{cases}$

Analysing the irrational equation, we get $\dfrac{2a}{c} = 1 \leftrightarrow c = 2a$ . By plugging this into the rational equation, we yield $\dfrac{a^2+b}{4a^2} = a \leftrightarrow b = a^2 (4a-1)$ .

But because $b$ is square free, we ought to have $a = 1$ . Thus, we get $b = 3$ and $c = 2$ .

Our desired answer is $\boxed{6.}$