jmc

Let $y=\frac{x}{\sqrt{2}}.$ Then $x=y\sqrt{2}.$ Substituting, we get $$18y^3\sqrt{2}-20y\sqrt{2}=8\sqrt{2},$$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 $a+b+c=2+14+3=\boxed{19}.$