Is this Vieta's Formula?

$\dfrac{α^2+α+1}{α^2-α+1}=\dfrac{α^3-1}{(α-1)(α^2-α+1)}$ . Since $α$ is a root of the equation $x^3-2x^2+6x-1=0$ , therefore $α^3-1=2α(α-3)$ and $(α-1)(α^2-α+1)=-4α$ . So $α\dfrac{α^2+α+1}{α^2-α+1}=\dfrac{3α-α^2}{2}$ . Similar for $β$ and $\gamma$ . Therefore the given expression equals $\dfrac{3(α+β+\gamma) -(α^2+β^2+\gamma^2)}{2}$ . Now $α+β+\gamma=2, αβ+β\gamma+α\gamma=6$ . So, $α^2+β^2+\gamma^2=-8$ , and the value of the given expression is $\dfrac{3\times 2+8}{2}=\boxed 7$ . (Here $α=0.17609,β=0.91195+2.20163i$ and $\gamma=0.91195-2.20163i$ )

From the given equation:

$\begin{aligned} x^3 - 2x^2 + 6x - 1 & = 0 \\ x^3 - x^2 + 5x & = x^2 - x + 1 \\ x(x^2 - x+1) + 4x & = x^2 - x +1 \\ \implies x^2 - x +1 & = \frac {4x}{1-x} \\ \implies x \left(\frac {x^2+x+1}{x^2 - x+1} \right) & = x \left(\frac {(x^2+x+1)(1-x)}{4x} \right) = \frac {(x^2+x+1)(1-x)}4 = \frac {1-x^3}4 \end{aligned}$

Therefore,

$\begin{aligned} S & = \alpha \left(\frac {\alpha^2+\alpha+1}{\alpha^2 - \alpha+1} \right) + \beta \left(\frac {\beta^2+\beta+1}{\beta^2-\beta+1} \right) + \gamma \left(\frac {\gamma^2+\gamma+1}{\gamma^2-\gamma+1}\right) \\ & = \frac {1-\alpha^3}4 + \frac {1-\beta^3}4 + \frac {1-\gamma^3}4 = \frac 34 - \frac {\alpha^3+\beta^3+\gamma^3}4 \\ & = \frac 34 - \frac {(\alpha+\beta+\gamma)\left((\alpha+\beta+\gamma)^2-3(\alpha\beta+\beta\gamma + \gamma\alpha)\right) + 3\alpha\beta\gamma}4 & \small \blue{\text{By Vieta's formula}} \\ & = \frac 34 - \frac {2\left(2^2-3(6)\right) + 3(1)}4 = \frac 34 + \frac {25}4 = \boxed 7 \end{aligned}$

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Reference: Vieta's formula