jmc

In general, the solutions to cubic equations are very messy, so we hope there is a trick to solving this particular equation.

Noticing the $(3,3,1)$ pattern of coefficients, which appears in the expansion $$(x+1{)}^3=x^3+3x^2+3x+1,$$we rewrite the left-hand side as $$9x^3-(x^3+3x^2+3x+1)=0$$or $$9x^3-(x+1{)}^3=0.$$Thus, $9x^3=(x+1{)}^3$, and since $x$ is real, $$x\sqrt[3]{9}=x+1⟹x=\frac{1}{\sqrt[3]{9}-1}.$$To rationalize the denominator, we write $$x=\frac{1}{\sqrt[3]{9}-1}\cdot \frac{\sqrt[3]{81}+\sqrt[3]{9}+1}{\sqrt[3]{81}+\sqrt[3]{9}+1}=\frac{\sqrt[3]{81}+\sqrt[3]{9}+1}{8}$$by the difference of cubes factorization. The answer is $81+9+8=\boxed{98}$.