A) \[{{x}^{3}}-4x-1=0\]
B) \[{{x}^{3}}-4x+1=0\]
C) \[{{x}^{3}}+4x-1=0\]
D) \[{{x}^{3}}+4x+1=0\]
Correct Answer: C
Solution :
Given,\[\alpha \],\[\beta \]and\[\gamma \]are the roots of\[{{x}^{3}}+4x+1=0\]. \[\therefore \]\[\alpha +\beta +\gamma =0,\,\,\alpha \beta +\beta \gamma +\gamma \alpha =4,\,\,\alpha \beta \gamma =-1\] Now,\[\frac{{{\alpha }^{2}}}{\beta +\gamma }+\frac{{{\beta }^{2}}}{\gamma +\alpha }+\frac{{{\gamma }^{2}}}{\alpha +\beta }=\frac{{{\alpha }^{2}}}{-\alpha }+\frac{{{\beta }^{2}}}{-\beta }+\frac{{{\gamma }^{2}}}{\gamma }\] \[=-(\alpha +\beta +\gamma )=0\] \[\frac{{{\alpha }^{2}}{{\beta }^{2}}}{(\beta +\gamma )(\gamma +\alpha )}+\frac{{{\beta }^{2}}{{\gamma }^{2}}}{(\gamma +\alpha )(\alpha +\beta )}+\frac{{{\gamma }^{2}}{{\alpha }^{2}}}{(\beta +\gamma )(\alpha +\beta )}\] \[=\alpha \beta +\beta \gamma +\gamma \alpha =4\] and\[\frac{{{\alpha }^{2}}{{\beta }^{2}}{{\gamma }^{2}}}{(\beta +\gamma )(\gamma +\alpha )(\alpha +\beta )}=-\alpha \beta \gamma =1\] \[(\because \,\,\alpha +\beta +\gamma =0)\] \[\therefore \]Required equation is \[{{x}^{3}}+4x-1=0\]
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