If m is chosen in the quadratic equation ( m ^ 2 + 1) x ^ 2 - 3x +...

Let, the roots of $\left({m}^{2} + 1\right){x}^{2} - 3x + {\left({m}^{2} + 1\right)}^{2} = 0$ are $\alpha$ and $\beta$

We know that the sum and product of roots of a quadratic equation $a{x}^{2} + bx + c = 0, a \ne 0$ are respectively $- \dfrac{b}{a}$ and $\dfrac{c}{a} .$

$∴ \alpha + \beta = \dfrac{3}{{m}^{2} + 1}$ and $\alpha\beta = \dfrac{{\left({m}^{2} + 1\right)}^{2}}{\left({m}^{2} + 1\right)} = \left({m}^{2} + 1\right)$

$∵ \alpha + \beta$ is maximum, $∴ m = 0$

$∴$ the equation is ${x}^{2} - 3x + 1 = 0$

And $\alpha + \beta = 3, \alpha\beta = 1$

Now, ${\alpha}^{3} - {\beta}^{3} = \left(\alpha - \beta\right)\left({\alpha}^{2} + {\beta}^{2} + \alpha\beta\right)$

$= \sqrt{{\left(\alpha + \beta\right)}^{2} - 4\alpha\beta}\left({\left(\alpha + \beta\right)}^{2} - \alpha\beta\right)$

$= \sqrt{9 - 4}\left(9 - 1\right)$

$= 8\sqrt{5} .$