A) \[-\frac{1}{2}\]
B) \[\frac{1}{2}\]
C) 32
D) \[\frac{1}{4}\]
Correct Answer: D
Solution :
Here, \[\alpha +\beta =-2\] and \[\alpha \beta =4\] \[\therefore \frac{1}{{{\alpha }^{3}}}+\frac{1}{{{\beta }^{3}}}=\frac{{{\alpha }^{3}}+{{\beta }^{3}}}{{{(\alpha \beta )}^{3}}}\]\[=\frac{{{(\alpha +\beta )}^{3}}-3\alpha \beta (\alpha +\beta )}{{{(\alpha \beta )}^{3}}}\] \[=\frac{{{(-2)}^{3}}-3(-2)(4)}{{{(4)}^{3}}}\] =\[\frac{16}{64}=\frac{1}{4}\].
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