If m is chosen in the quadratic equation (m² + 1)x² - 3x + (m² + 1)² = 0 such that the sum of its roots is greatest, then the absolute difference of the cubes of its roots is

Given quadratic equation is
$\left({\mathrm{m}}^2+1\right){\mathrm{x}}^2-3\mathrm{x}+{\left({\mathrm{m}}^2+1\right)}^2=0...\left(\mathrm{i}\right)$
Let the roots of quadratic Eq. (i) are $\mathrm{\alpha }$, and $\mathrm{\beta }$, so
$\mathrm{\alpha }+\mathrm{\beta }=\frac{3}{{\mathrm{m}}^2+1}\text{ and }\mathrm{\alpha \beta }={\mathrm{m}}^2+1$
According to the question, the sum of roots is greatest and it is possible only when "(m² + 1) is min" and "min value of m² + 1=1, when m =0".
$\therefore \mathrm{\alpha }+\mathrm{\beta }=3\text{ and }\mathrm{\alpha \beta }=1,\text{ as }\mathrm{m}=0$
Now, the absolute difference of the cubes of root
$\begin{array}{l}=\left|{\mathrm{\alpha }}^3-{\mathrm{\beta }}^3\right|=|\mathrm{\alpha }-\mathrm{\beta }|\left|{\mathrm{\alpha }}^2+{\mathrm{\beta }}^2+\mathrm{\alpha \beta }\right|\\ =\sqrt{(\mathrm{\alpha }+\mathrm{\beta }{)}^2-4\mathrm{\alpha \beta }} \left|(\mathrm{\alpha }+\mathrm{\beta }{)}^2-\mathrm{\alpha \beta }\right|\\ =\sqrt{9-4}|9-1|=8\sqrt{5}\end{array}$