A) \[\frac{c(a-b)}{{{a}^{2}}}\]
B) 0
C) \[-\frac{bc}{{{a}^{2}}}\]
D) None of these
Correct Answer: A
Solution :
\[\alpha +\beta =-\frac{b}{a}\]and\[\alpha \beta =\frac{c}{a}\] Now \[\alpha {{\beta }^{2}}+{{\alpha }^{2}}\beta +\alpha \beta =\alpha \beta (\beta +\alpha )+\alpha \beta \] \[=\alpha \beta (1+\alpha +\beta )=\frac{c}{a}\left\{ 1+\left( -\frac{b}{a} \right) \right\}=\frac{c(a-b)}{{{a}^{2}}}\]
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