A) \[\frac{3ABC-{{B}^{3}}}{{{A}^{3}}}\]
B) \[\frac{3ABC+{{B}^{3}}}{{{A}^{3}}}\]
C) \[\frac{{{B}^{3}}-3ABC}{{{A}^{3}}}\]
D) \[\frac{{{B}^{3}}-3ABC}{{{B}^{3}}}\]
Correct Answer: A
Solution :
Given equation, \[A{{x}^{2}}+Bx+C=0\] Þ \[\alpha +\beta =-\frac{B}{A},\alpha \beta =\frac{C}{A}\] \ \[{{\alpha }^{3}}+{{\beta }^{3}}={{(\alpha +\beta )}^{3}}-3\alpha \beta (\alpha +\beta )\] \[={{\left( -\frac{B}{A} \right)}^{3}}-3\left( \frac{C}{A} \right)\,\left( -\frac{B}{A} \right)\]\[=-\frac{{{B}^{3}}}{{{A}^{3}}}+\frac{3BC}{{{A}^{2}}}\] \[{{\alpha }^{3}}+{{\beta }^{3}}=\frac{3ABC-{{B}^{3}}}{{{A}^{3}}}\].
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