If λ ∈ R is such that the sum of the cubes of the roots of the...

{x}^{2} + \left(2 - \lambda\right)x + \left(10 - \lambda\right) = 0$</p><p>$⇒ \alpha + \beta = \lambda - 2 \text{$and$} \alpha\beta = 10 - \lambda . . . . . . . . . . \left(i\right)$</p><p>Let roots are$\alpha and\beta$.</p><p>$⇒ {\alpha}^{3} + {\beta}^{3} = {\left(\alpha + \beta\right)}^{3} - 3\alpha\beta ( \alpha + \beta )$</p><p>$= {\left(\lambda - 2\right)}^{3} - 3\left(10 - \lambda\right) ( \lambda - 2 )$</p><p>$= {\lambda}^{3} - {6\lambda}^{2} + 12\lambda - 8 - 3 ( 10\lambda - {\lambda}^{2} - 20 + 2\lambda )$</p><p>$= {\lambda}^{3} - {3\lambda}^{2} - 24\lambda + 52$</p><p>$\dfrac{dz}{d\lambda} = 3{\lambda}^{2} - 6\lambda - 24 = 3$\left({\lambda}^{2} - 2\lambda - 8\right)$ (where, $z = {\alpha}^{3} + {\beta}^{3}$)

For maximum and critical points, derivative must be zero.

$⇒ {\lambda}^{2} - 2\lambda - 8 = 0$

$⇒ \left(\lambda - 4\right)\left(\lambda + 2\right) = 0$

$⇒ \lambda = - 2,4$

Now, $\dfrac{{d}^{2}z}{d{\lambda}^{2}} = 6\lambda - 6$

For \left(\lambda = - 2\right), \dfrac{{d}^{2}z}{d{\lambda}^{2}} < 0 ⇒ {\alpha}^{3} + {\beta}^{3}$is maximum and</p><p>For$( \lambda = 4 ) , \dfrac{{d}^{2}z}{d{\lambda}^{2}} > 0 ⇒ {\alpha}^{3} + {\beta}^{3}$is minimum.</p><p>$⇒$Equation will be${x}^{2} - 2x + 6 = 0$.</p><p>Using quadratic formula, we get</p><p>$x = \dfrac{2 \pm \sqrt{{\left(- 2\right)}^{2} - 4 \times 1 \times 6}}{2 \times 1} = \dfrac{2 \pm \sqrt{- 20}}{2} = \dfrac{2 \pm 2\sqrt{5}i}{2} = 1 \pm \sqrt{5}i$</p><p>Thus, difference of roots is \left|\alpha - \beta\right| = 2\sqrt{5}$