| If \[x+\frac{1}{x}=\sqrt{3},\] then the value of \[{{x}^{18}}+{{x}^{12}}+{{x}^{6}}+1\]is [SSC (CPO) 2011] |
A) 0
B) 1
C) 2
D) 8
Correct Answer: A
Solution :
| Given, \[x+\frac{1}{x}=\sqrt{3}\] (i) |
| Cubing on both sides, we get |
| \[{{\left( x+\frac{1}{x} \right)}^{3}}={{(\sqrt{3})}^{3}}\] |
| \[[\because {{(a+b)}^{3}}={{a}^{3}}+{{b}^{3}}+3ab(a+b)]\] |
| \[\Rightarrow \]\[{{x}^{3}}+\frac{1}{{{x}^{3}}}+3\left( x+\frac{1}{x} \right)={{(\sqrt{3})}^{3}}\] [from Eq.(i)] |
| \[\Rightarrow \]\[{{x}^{3}}+\frac{1}{{{x}^{3}}}+3\sqrt{3}=3\sqrt{3}\]\[\Rightarrow \]\[{{x}^{3}}+\frac{1}{{{x}^{3}}}=0\] |
| \[\Rightarrow \]\[{{x}^{16}}+{{x}^{12}}+{{x}^{6}}+1={{x}^{12}}({{x}^{6}}+1)+1({{x}^{2}}+1)\] |
| \[=({{x}^{12}}+1)({{x}^{6}}+1)\] |
| \[=({{x}^{12}}+1)\cdot {{x}^{3}}\left( {{x}^{3}}+\frac{1}{{{x}^{3}}} \right)=0\] |
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