jmc

We can arrange the equation as $$9x^3=x^3+3x^2+3x+1=(x+1{)}^3.$$Taking the cube root of both sides, we get $$x\sqrt[3]{9}=x+1.$$Then $(\sqrt[3]{9}-1)x=1$, so $$x=\frac{1}{\sqrt[3]{9}-1}.$$To rationalize the denominator, we multiply the numerator and denominator by $\sqrt[3]{9^2}+\sqrt[3]{9}+1.$ This gives us $$\frac{\sqrt[3]{9^2}+\sqrt[3]{9}+1}{(\sqrt[3]{9}-1)(\sqrt[3]{9^2}+\sqrt[3]{9}+1)}=\frac{\sqrt[3]{81}+\sqrt[3]{9}+1}{8}.$$Then $a+b+c=81+9+8=\boxed{98}.$