Let $ y = \frac{x}{\sqrt{2}} $. Then $ x = y \sqrt{2} $. Substituting, we get \[18 y^3 \sqrt{2} - 20y \sqrt{2} = 8 \sqrt{2},\]so $ 18y^3 - 20y - 8 = 0 $. Dividing by 2, we get $ 9y^3 - 10y - 4 = 0 $. Looking for rational roots, we find $ y = -\frac{2}{3} $ works. Thus, we can take out a factor of $ 3y + 2, $ to get \[(3y + 2)(3y^2 - 2y - 2) = 0.\]The roots of $ 3y^2 - 2y - 2 = 0 $ are $ \frac{1 \pm \sqrt{7}}{3} $. Therefore, the solutions $ x $ are $ -\frac{2 \sqrt{2}}{3} $ and $ \frac{\sqrt{2} \pm \sqrt{14}}{3} $. The largest solution is $ \frac{\sqrt{2} + \sqrt{14}}{3}, $ so $ a + b + c = 2 + 14 + 3 = \boxed{19} $.