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 ism2+1x2−3x+m2+12=0...(i)Let the roots of quadratic Eq. (i) are α, and β, soα+β=3m2+1 and αβ=m2+1According 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".∴α+β=3 and αβ=1, as m=0Now, the absolute difference of the cubes of root=α3−β3=|α−β|α2+β2+αβ=(α+β)2−4αβ (α+β)2−αβ=9−4|9−1|=85