A) \[-\frac{1}{2}\]
B) \[\frac{1}{2}\]
C) \[32\]
D) \[\frac{1}{32}\]
E) \[\frac{1}{4}\]
Correct Answer: E
Solution :
\[\because \] \[\alpha \]and\[\beta \]are the roots of equation \[{{x}^{2}}+2x+4=0\] \[\therefore \] \[\alpha +\beta =-2,\alpha \beta =4\] \[\therefore \] \[\frac{1}{{{\alpha }^{3}}}+\frac{1}{{{\beta }^{3}}}=\frac{({{\alpha }^{3}}+{{\beta }^{3}})}{{{(\alpha \beta )}^{3}}}\] \[=\frac{{{(\alpha +\beta )}^{3}}-3\alpha \beta (\alpha +\beta )}{{{(\alpha \beta )}^{3}}}\] \[=\frac{-8+24}{64}=\frac{1}{4}\]
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