$\because(3+\sqrt{2})^6-(3-\sqrt{2})^6=a+b \sqrt{2}$
...(i)
$\begin{aligned} & \text { Let } x=3+\sqrt{2} \text { and } y=3-\sqrt{2} \\ & x^2+y^2=9+2+6 \sqrt{2}+9+2-6 \sqrt{2}=22 \\ & x^3+y^3=(3+\sqrt{2})^3+(3-\sqrt{2})^3\end{aligned}$
$\begin{aligned}=3^3+(\sqrt{2})^3+3 \cdot 3 \cdot \sqrt{2}(3 & +\sqrt{2}) \\ & +3^3-(\sqrt{2})^3-3 \cdot 3 \cdot \sqrt{2}(3-\sqrt{2})\end{aligned}$
$\begin{aligned} & =54+36=90 \\ & (3+\sqrt{2})^6-(3-\sqrt{2})^6=x^6-y^6 \\ & =\left(x^3+y^3\right)\left(x^3-y^3\right) \\ & =90(x-y)\left(x^2+y^2+x y\right) \\ & =90.2 \sqrt{2}(22+9-2) \\ & =58 \times 90 \sqrt{2}=5220 \sqrt{2} \\ & \therefore b=5220 \text { and } a=0 \\ & a+b=0+5220=5220 .\end{aligned}$