If m is chosen in the quadratic equation m 2 + 1 x 2 - 3 x + m 2 + 1 2 = 0 such that the sum of its roots is…

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 $-\frac{b}{a}$ and $\frac{c}{a}.$ 

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

$\because \alpha +\beta$ is maximum, $\therefore m=0$

$\therefore$ the equation is $x^2-3x+1=0$

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

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

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

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

$=8\sqrt{5}.$