Newton's Sums, here?

Using Vieta's formulas, we have:

$\begin{cases} a + b + c = -2 \\ ab + bc +ca = -3 \\ abc = 1 \end{cases}$

$\Rightarrow \begin{cases} \dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c} = \dfrac {ab + bc +ca}{abc} = -3 \\ \dfrac{1}{ab} + \dfrac{1}{bc} + \dfrac{1}{ca} = \dfrac {a + b +c}{abc} = -2 \\ \dfrac{1}{abc} = 1 \end{cases}$

$\begin{aligned} \sum_{cyc} \frac{1}{a^3} & = \dfrac{1}{a^3} + \dfrac{1}{b^3} + \dfrac{1}{c^3} \\ & = \left( \dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c} \right) \left( \dfrac{1}{a^2} + \dfrac{1}{b^2} + \dfrac{1}{c^2} \right) \\ & \quad - \left( \dfrac{1}{ab} + \dfrac{1}{bc} + \dfrac{1}{ca} \right) \left( \dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c} \right) + 3 \left( \dfrac{1}{abc} \right) \\ & = \left( \dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c} \right) \left( \left[\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c} \right]^2 - 2\left[ \dfrac{1}{ab} + \dfrac{1}{bc} + \dfrac{1}{ca}\right] \right) \\ & \quad - \left( \dfrac{1}{ab} + \dfrac{1}{bc} + \dfrac{1}{ca} \right) \left( \dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c} \right) + 3 \left( \dfrac{1}{abc} \right) \\ & = \left( \dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c} \right)^3 - 3 \left( \dfrac{1}{ab} + \dfrac{1}{bc} + \dfrac{1}{ca} \right) \left( \dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c} \right) + 3 \left( \dfrac{1}{abc} \right) \\ & = (-3)^3 - 3(-2)(-3) + 3(1) = -27 - 18 + 3 = \boxed{-42} \end{aligned}$

Let us consider the transformation, $x = \dfrac{1}{y}$ .

$\dfrac{1}{y^3} + \dfrac{2}{y^2} - \dfrac{3}{y} - 1 = 0$

$\Rightarrow y^3 + 3y^2 - 2y - 1 = 0$

Let the roots of this equation be $d , e , \text{ \& } f$ .

$S = \displaystyle \sum_{\text{cyc}} \dfrac{1}{a^3} = d^3 + e^3 + f^3$

Defining $S_1 = d + e + f$ , $S_2 = d^2 + e^2 + f^2$ and $S_3 = d^3 + e^3 + f^3$ ,

By Newton's Sums,

$1.S_1 + 3 = 0 \Rightarrow S_1 = - 3$

$1.S_2 + 3.S_1 + 2(-2) = 0 \Rightarrow S_2 = 13$

$1.S_3 + 3.S_2 + (-2).S_1 + 3(-1) = 0 \Rightarrow S_3 = \boxed{-42}$

Hence,

$S = \sum_{\text{cyc}} \dfrac{1}{a^3} = d^3 + e^3 + f^3 = S_3 = \color{#D61F06}{\boxed{-42}}$

---

Note:

• $a.b$ denotes multiplication.

Given equation is :

$x^3+2x^2-3x-1=0 \implies 1+\dfrac{2}{x}-\dfrac{3}{x^2}=\dfrac{1}{x^3}$

Also, $\dfrac{1}{x^2}=x+2-\dfrac{3}{x}$

Hence ,

$\dfrac{1}{x^3}=-3x-5+\dfrac{11}{x}$

So, the required sum is equivalent to the following :

$-3(a+b+c)-3(5)+11\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)$

The above expression is easy to evaluate and gives answer as $-42$

$1/a+1/b+1/c= - (-3)/(-1)= - 3.\\ 1/ab+1/bc+1/ca=+2/(-1)= - 2.\\ 1/abc = - 1/(-1)= 1.\\ \therefore~(1/a+1/b+1/c)^2=(1/a^2+1/b^2+1/c^2)+2(1/ab+1/bc+1/ca),\\ \implies~(1/a^2+1/b^2+1/c^2)= 9-2(-2)=13.\\ 1/a^3+1/b^3+1/c^3= 3/abc+\{ 1/a+1/b+1/c\}\{(1/a^2+1/b^2+1/c^2) - (1/ab+1/bc+1/ca)\},\\ = 3 +\{-3\}\{13 - (-2)\} \\ = \Large \color{#D61F06}{ - 42}.$