Consider the quadratic equation (n^2 - 2n + 2)x^2 - 3x + (n^2 - 2n...

The given quadratic equation is $(n^2 - 2n + 2)x^2 - 3x + (n^2 - 2n + 2)^2 = 0$.

Let $k = n^2 - 2n + 2 = (n-1)^2 + 1$. Since $(n-1)^2 \ge 0$ for all $n \in \mathbb{R}$, the minimum value of $k$ is $1$ at $n=1$.

The product of the roots is given by $P = \dfrac{(n^2 - 2n + 2)^2}{n^2 - 2n + 2} = n^2 - 2n + 2 = k$.

The minimum value of the product of the roots is $\alpha = \min(k) = 1$.

The sum of the roots is given by $S = \dfrac{3}{n^2 - 2n + 2} = \dfrac{3}{k}$.

The maximum value of the sum of the roots occurs when $k$ is minimum. Thus, $\beta = \max\left(\dfrac{3}{k}\right) = \dfrac{3}{1} = 3$.

We are given a Geometric Progression (G.P.) whose first term is $a = \alpha = 1$ and common ratio is $r = \dfrac{\alpha}{\beta} = \dfrac{1}{3}$.

The sum of the first six terms of this G.P. is:

$S_6 = a \dfrac{1 - r^6}{1 - r} = 1 \cdot \dfrac{1 - \left(\dfrac{1}{3}\right)^6}{1 - \dfrac{1}{3}}$

$S_6 = \dfrac{1 - \dfrac{1}{729}}{\dfrac{2}{3}} = \dfrac{\dfrac{728}{729}}{\dfrac{2}{3}}$

$S_6 = \dfrac{728}{729} \times \dfrac{3}{2} = \dfrac{364}{243}$

Answer: $\dfrac{364}{243}$