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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: $$A=98$$

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  • $\begingroup$ You applied Vieta's formulas in the wrong way, since if $$ f(x)=a_0 +a_1 x + a_2 x^2 + a_3 x^3 $$ and the roots of $f$ are $\zeta_1,\zeta_2,\zeta_3$, we have: $$\zeta_1\zeta_2\zeta_3 = -\frac{a_0}{a_3},\qquad \zeta_1+\zeta_2+\zeta_3 = -\frac{a_2}{a_3}.$$ By the way, Vieta's formulas are not the fastest option in this case, just see below. $\endgroup$ – Jack D'Aurizio Mar 3 '15 at 14:47
  • $\begingroup$ Note that your problem does not ask the sum of the roots of the polynomial! The problem is very different: you have to compute only the real root of $f(x)$, ignoring the other two roots. $\endgroup$ – Crostul Mar 3 '15 at 14:48
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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.

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