$c+b = ?$

Given that $a^2 - 3^\frac 23 - 3^{-\frac 23} + 2 = 0$ , therefore,

$\begin{aligned} a^2 & = 3^\frac 23 - 2 + 3^{-\frac 23} \\ & = \left(\sqrt[3] 3-\frac 1{\sqrt[3]3}\right)^2 \\ \implies a & = \sqrt[3] 3-\frac 1{\sqrt[3]3} & \small \color{#3D99F6} \text{Taking }a > 0 \\ a^3 & = 3 - 3\sqrt[3]3+\frac 3{\sqrt[3]3} - \frac 13 \\ & = \frac 83 - 3\left(\sqrt[3]3-\frac 1{\sqrt[3]3}\right) \\ & = \frac 83 - 3a \\ \implies 9a^3 & = 24 - 27a \end{aligned}$

Therefore, $b+c=24+27 = \boxed{51}$ .

$\large \begin{aligned} a^2 - 3^{\frac 23} - 3^{-\frac 23} + 2 & = 0 \\ \Rightarrow a^2 & = 3^{\frac23} + 3^{-\frac 23} -2 \\ \Rightarrow a^2 & = \left(3^{\frac 13} \right)^2 + \left(3^{\frac 13} \right)^2 - 2 \cdot 3^{\frac 13} \cdot 3^{-\frac13} \\ \Rightarrow a^2 & = \left(3^{\frac 13} - 3^{-\frac 13} \right)^2 \\ \Rightarrow a & = 3^{\frac 13} - 3^{-\frac 13} \\ \Rightarrow a^3 & = \left(3^{\frac 13} - 3^{-\frac 13} \right)^3 ~~~~~[\text{Cube both sides}] \\ \Rightarrow a^3 & = \left(3^{\frac 13}\right)^3 - \left(3^{-\frac 13} \right)^3 - 3 \cdot 3^{\frac 13} \cdot 3^{-\frac 13} \left(3^{\frac 13} - 3^{-\frac 13} \right) \\ \Rightarrow a^3 & = 3 -3^{-1} - 3 \cdot 3^0 a ~~~~ \left[3^{\frac 13} - 3^{-\frac 13} = a \right] \\ \Rightarrow a^3 & = 3 - \frac 13 - 3a \\ \Rightarrow 9a^3 & = 27 - 3 - 3(9)a ~~~~[\text{Multiply both sides by 9}] \\ \implies 9a^3 & = 24 - 27a \\ \end{aligned}$

So $c = 24$ and $b = 27$ . Hence $b + c = 24 + 27 = \boxed{51}$

Note: $(a - b)^3 = a^3 - b^3 - 3ab(a-b)$