Sum of the roots of an equation. 
The real root of the equation $8x^3 - 3x^2 - 3x - 1 = 0$ can be written in the form $\frac{\sqrt[3]a + \sqrt[3]b + 1}{c}$, where $a$, $b$, and $c$ are positive integers. Find $a+b+c$.

I used Vieta's to get the wrong answer. 
If $$f(x) = a_0 + a_1x + a_2x^2 + a_3x^3$$
And $p, b, c$ are the roots then: (here let $a = a_3$ the high degree coefficient) 
$$f(x) = a(x - p)(x - b)(x - c) = ax^3 - apx^2 - acx^2 + acpx - abx^2 + abpx + abcx - abcp$$ 
$$ = ax^2 + x^2\overbrace{(-ap - ac - ab)}^{a_2} + x\overbrace{(acp + abp + abc)}^{a_1} - abcp$$
$$f(x) = 8x^3 - 3x^2 - 3x - 1$$
We are after $a_2$
$$-3 = -(ap + ac + ab) \implies p + c + b = 3/a = \frac{3}{8}$$
Which is wrong because, the answer is supposed to be an integer, and the answer is: $$A=98$$

We have:
$$ p(x)=8x^3-3x^2-3x-1 = 9x^3-(x+1)^3 $$
hence $p(x)=0$ iff
$$ \left(1+\frac{1}{x}\right)^3 = 9 $$
and the only real root of the above equation is given by
$$ x = \frac{1+\sqrt[3]{81}+\sqrt[3]{9}}{8} $$
leading to the correct answer.
