| If \[x=3+2\sqrt{2},\] then the values of \[{{x}^{3}}+\frac{1}{{{x}^{3}}}\] and \[{{x}^{3}}-\frac{1}{{{x}^{3}}}\] are respectively. |
A) \[140\sqrt{2},\] 198
B) 234, 216
C) 216, 234
D) 198, \[140\sqrt{2}\]
Correct Answer: D
Solution :
| \[x=3+2\sqrt{2}\] |
| \[x+\frac{1}{x}=3+2\sqrt{2}+\frac{1}{3+2\sqrt{2}}\times \frac{(3-2\sqrt{2})}{(3-2\sqrt{2})}\] \[=3+2\sqrt{2}+\frac{3-2\sqrt{2}}{9-8}=3+2\sqrt{2}+3-2\sqrt{2}\] |
| \[\Rightarrow \] \[x+\frac{1}{x}=6\] |
| On cubing both sides, we get |
| \[{{x}^{3}}+\frac{1}{{{x}^{3}}}+3\left( x+\frac{1}{x} \right)x\times \frac{1}{x}={{6}^{3}}\] |
| \[\Rightarrow \] \[{{x}^{3}}+\frac{1}{{{x}^{3}}}+3\,(6)=216\] |
| \[\Rightarrow \] \[{{x}^{3}}+\frac{1}{{{x}^{3}}}=216-18=198\] |
| Similarly, \[{{x}^{3}}-\frac{1}{{{x}^{3}}}=140\sqrt{2}\] |
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