The sum of all the value of m for which the roots $x_{1} a n d x_{2}$ of the quadratic equation $x^{2} - 2 m x + m = 0$ satisfy the condition $x_{1}^{3} + x_{2}^{3} = x_{1}^{2} + x_{2}^{2}$ , is :

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Solution

$\begin{matrix} {x_{1}}^{3} + {x_{2}}^{3} = {x_{1}}^{2} + {x_{2}}^{2} (x_{1} + x_{2}) ({x_{1}}^{2} + {x_{2}}^{2} - 2 x_{1} x_{2}) = {(x_{1} + x_{2})}^{2} - 2 x_{1} x_{2} (2 m) (4 m^{2} - 3 m) = 4 m^{2} - 2 m 8 m^{3} - 6 m^{2} = 4 m^{2} - 2 m 8 m^{2} - 6 m = 4 m - 2 8 m^{2} - 10 m + 2 = 0 8 m^{2} - 8 m - 2 m + 2 = 0 8 m (m - 1) - 2 (m - 1) = 0 (8 m - 2) (m - 1) = 0 m = 1, \frac{1}{4} \end{matrix}$

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