A) 7
B) 12
C) 18
D) 36
Correct Answer: C
Solution :
Given, \[\alpha +\beta +\gamma =2,\,{{\alpha }^{2}}+{{\beta }^{2}}+{{\gamma }^{2}}=6,\]\[{{\alpha }^{3}}+{{\beta }^{3}}+{{\gamma }^{3}}=8\] Now, \[(\alpha +\beta +{{\gamma }^{2}})={{2}^{2}}\] \[\Rightarrow \]\[{{\alpha }^{2}}+{{\beta }^{2}}+{{\gamma }^{2}}+2(\alpha \beta +\beta \gamma +\gamma \alpha )=4\] \[\Rightarrow \]\[2(\alpha \beta +\beta \gamma +\gamma \alpha )=4-6=-2\] Also, \[{{\alpha }^{3}}+{{\beta }^{3}}+{{\gamma }^{3}}-3\alpha \beta \gamma \] \[=(\alpha +\,\beta +\gamma )({{\alpha }^{2}}+{{\beta }^{2}}+{{\gamma }^{2}}-\alpha \beta -\beta \gamma -\gamma \alpha )\] \[\Rightarrow \]\[8-3\alpha \beta \gamma =2[6-(-1)]\] \[\Rightarrow \]\[8-3\alpha \beta \gamma =14\] \[\Rightarrow \]\[\alpha \beta \gamma =8-14\] \[\Rightarrow \]\[\alpha \beta \gamma =-2\] Now, \[{{\alpha }^{4}}+{{\beta }^{4}}+{{\gamma }^{4}}={{({{\alpha }^{2}}+{{\beta }^{2}}+{{\gamma }^{2}})}^{2}}-2\sum {{\alpha }^{2}}{{\beta }^{2}}\] \[=({{\alpha }^{2}}+{{\beta }^{2}}+{{\gamma }^{2}})-2[{{(\sum \beta \gamma )}^{2}}-2\alpha \beta \gamma \sum \alpha ]\] \[={{(6)}^{2}}-2[{{(-1)}^{2}}-2(-2)2]\] \[=36-2[9]\] \[=36-18\] \[=18\]
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