Submitted by devansh shringi

We have that $9x^3 = (x+1)^3$ , so it follows that $\sqrt[3]{9}x = x+1$ . Solving for $x$ yields $\frac{1}{\sqrt[3]{9}-1} = \frac{\sqrt[3]{81}+\sqrt[3]{9}+1}{8}$ , so the answer is $\boxed{098}$ .

This user blatantly stole the problem from the 2013 AIME I problem 5. It seems fitting that I should, similarly, blatantly copy a solution that can be found here: http://www.artofproblemsolving.com/Wiki/index.php/2013 AIME I Problems/Problem 5

@devansh shringi: It is OK for you to get inspiration from other problems, but it is not OK to just copy them and submit them as your own.

I really liked the simplicity of this (apparently hard) problem.

Let's rewrite $f(x) = 8x^3 - 3x^2 - 3x - 1$ as $f(x) = 9x^3 - (x+1)^3$ .

Knowing $(a - b)(a^2 + ab + b^2)= a^3 - b^3$ , we can simplify $f(x)$ to $(\sqrt[3]{9}x - x - 1)(\sqrt[3]{81} +\sqrt[3]{9}(x-1) + (x-1)^2) = 0$ .

It is clear that the expression $\sqrt[3]{9}x - x - 1 = 0$ holds the real root. Thus $x = \dfrac{1}{\sqrt[3]{9} - 1}$ .

Remembering $(a - 1)(a^2 + a + 1)= x^3 - 1$ , we can use this trick to simplify $x$ 's fraction to $x = \dfrac { \sqrt[3]{81} + \sqrt[3]{9} + 1 }{8}$ . Thus $a + b + c = 98$ .

Let $y=x-\frac{1}{8}$ , then $8x^3-3x^2-3x-1=0 \iff x^3-\frac{3}{8}x^2-\frac{3}{8}x-\frac{1}{8}=0 \iff y^3-\frac{27}{64}y-\frac{90}{512}=0$ . Now if $y=u+v$ then $y^3-3uvy-(u^3+v^3)=0$ has root $y=u+v$ . I need $\frac{9}{64}=uv$ and $\frac{90}{512}=u^3+v^3$ . From these two equations, $u^6-\frac{90}{512}u^3+\frac{9^3}{64^3}=0$ so $u^3=\frac{1}{2} \left(\frac{90}{512}+\sqrt{\frac{90^2}{512^2}-4\frac{9^3}{64^3}} \right )$ and $v^3=\frac{1}{2} \left (\frac{90}{512}-\sqrt{\frac{90^2}{512^2}-4\frac{9^3}{64^3}} \right )$ Therefore $y=\sqrt[3]{\frac{1}{2} \left ( \frac{90}{512}+\sqrt{\frac{9^2 \cdot 10^2}{512^2}-4\frac{9^3}{64^3}}\right )}+\sqrt[3]{\frac{1}{2} \left (\frac{90}{512}-\sqrt{\frac{9^2 \cdot 10^2}{512^2}-4\frac{9^3}{64^3}} \right )}$ $y=\sqrt[3]{\frac{1}{2^{10}} \left (90+\sqrt{9^2(100-4 \cdot 9)} \right )}+\sqrt[3]{\frac{1}{2^{10}} \left (90-\sqrt{9^2(100-4 \cdot 9)}\right ) }$ $y=\sqrt[3]{\frac{1}{2^{10}}(90+72)}+\sqrt[3]{\frac{1}{2^{10}}(90-72)}$ $y=\frac{3+3^{\frac{2}{3}}}{8} \iff x=\frac{81^{\frac{1}{3}}+9^{\frac{1}{3}}+1}{8}$ $a+b+c=98$

By applying cardano's method ,we get the real root as: $x = \frac{3^{\frac{4}{3}}+3^{\frac{2}{3}}+1}{8}$ Hence, a = 81, b = 9, c = 8 $\Rightarrow a+b+c = 81 + 9 + 8 = \boxed{98}$

f(x) = 0 adding x cubed to both sides yields 9x^3 = x^3 +3x^2 +3x +1 = (x+1)^3 taking cube root of both sides and then solving for x and rationalizing the denominator, x = (cubert (81) + cubert (9) + 1)/8 so the answer is 98