# If m is chosen in the quadratic equation $\left ( m^{2}+1 \right )x^{2}-3x+\left ( m^{2}+1 \right )^{2}=0$ such that the sum of its roots is greatest, then the absolute difference of the cubes of its roots is  : Option 1) $10\sqrt{5}$ Option 2) $8\sqrt{3}$ Option 3) $8\sqrt{5}$ Option 4) $4\sqrt{3}$

$\\\ sum\:\:of\:\:root = \frac{-b}{a} \\\\\:=-\frac{-3}{m^{2}+1}=\frac{3}{(m^{2}+1)}$

for sum of root to be greatest   $m^{2}+1$  should be minimum $\neq0$

$\\f^{'}(m)=m^{2}+1=0\\\\\:f^{'}(m)=2m=0$

$\therefore m=0$

now equation

$=x^{2}-3x+1=0$

$roots=\frac{3\pm \sqrt{9-4}}{2}= \frac{3\pm \sqrt{5}}{2}$

$\\\alpha+\beta=\frac{3}{m^{2}+1}=3\\\\\:\alpha\beta=\frac{(m+1)^{2}}{m^{2}+1}=1$

$\\\left | \alpha^{3}-\beta^{3} \right |=\left | (\alpha-\beta)(\alpha^{2}+\beta^{2}+\alpha\beta) \right |\\\\\:\left | \sqrt{(\alpha+\beta)^{2}-4\alpha\beta} \right | \left | (\alpha+\beta)^{2}-\alpha\beta \right |$

$=\sqrt{5}.8=8\sqrt{5}$

Option 1)

$10\sqrt{5}$

Option 2)

$8\sqrt{3}$

Option 3)

$8\sqrt{5}$

Option 4)

$4\sqrt{3}$

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