| If \[x=3+2\sqrt{2},\]then \[\frac{{{x}^{6}}+{{x}^{4}}+{{x}^{2}}+1}{{{x}^{3}}}\] is equal to |
A) 216
B) 192
C) 198
D) 204
Correct Answer: D
Solution :
| Given, \[x=3+2\sqrt{2}=3+\sqrt{8}\] |
| and \[\frac{1}{x}=\frac{1}{3+\sqrt{8}}\times \frac{3-\sqrt{8}}{3-\sqrt{8}}\] |
| \[=\frac{3-\sqrt{8}}{9-8}=(3-\sqrt{8})\] |
| Now, \[\frac{{{x}^{6}}+{{x}^{4}}+{{x}^{2}}+1}{{{x}^{3}}}=\frac{{{x}^{3}}\left( {{x}^{3}}+x+\frac{1}{x}+\frac{1}{{{x}^{3}}} \right)}{{{x}^{3}}}\] |
| \[=x\,\,({{x}^{2}}+1)+\frac{1}{x}\left( \frac{1}{{{x}^{2}}}+1 \right)\] (i) |
| Now, putting the values of x and \[\frac{1}{x}\] in Eq. (i), we get |
| \[=(3+\sqrt{8})\times [{{(3+\sqrt{8})}^{2}}+1]\] |
| \[+\,\,(3-\sqrt{8})[{{(3-\sqrt{8})}^{2}}+1]\] |
| \[=(3+\sqrt{8})(9+8+6\sqrt{8}+1)\] |
| \[+\,\,(3-\sqrt{8})(9+8-6\sqrt{8}+1)\] |
| \[=(3+\sqrt{8})(18+6\sqrt{8})+(3-\sqrt{8})\,\,(18-6\sqrt{8})\] |
| \[=(3+\sqrt{8})\,\,6\,(3+\sqrt{8})+(3-\sqrt{8})\,\,6\,\,(3-\sqrt{8})\] |
| \[=6{{(3+\sqrt{8})}^{2}}+6{{(3-\sqrt{8})}^{2}}\] |
| \[=6\,\,[{{(3+\sqrt{8})}^{2}}+{{(3-\sqrt{8})}^{2}}]\] |
| \[=6\times [9+8+6\sqrt{8}+9+8-6\sqrt{8}]\] |
| \[=6\times 34=204\] |
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