Solution:

Rewrite the equation as $9x^3 = x^3 + 3x^2 + 3x + 1 = (x + 1)^3.$ Taking real cube roots, $\sqrt[3]{9}\,x = x + 1,$ so $$x = \frac{1}{\sqrt[3]{9} - 1}.$$

Multiply numerator and denominator by $\sqrt[3]{81} + \sqrt[3]{9} + 1;$ the denominator becomes $(\sqrt[3]{9})^3 - 1 = 8,$ so $$x = \frac{\sqrt[3]{81} + \sqrt[3]{9} + 1}{8}.$$ Thus $a + b + c = 81 + 9 + 8 = 98.$