| \[x=3+2\sqrt{2},\]then the values of \[{{x}^{3}}+\frac{1}{{{x}^{3}}}\] and \[{{x}^{3}}-\frac{1}{{{x}^{3}}}\]are respectively [SSC (CGL) 2014] |
A) \[140\sqrt{2},\]\[198\]
B) \[234,\]\[216\]
C) \[216,\]\[234\]
D) \[198,\]\[140\sqrt{2}\]
Correct Answer: D
Solution :
| Given, \[x=3+2\sqrt{2}\] |
| \[\frac{1}{x}=\frac{1}{3+2\sqrt{2}}\times \frac{3-2\sqrt{2}}{3-2\sqrt{2}}\] |
| \[=3-2\sqrt{2}\] [on rationalising] |
| \[\therefore \]\[{{x}^{3}}+\frac{1}{{{x}^{3}}}={{(3+2\sqrt{2})}^{3}}+{{(3-2\sqrt{2})}^{3}}\] |
| \[=27+16\sqrt{2}+3\times 9\times 2\sqrt{2}+3\times 3\times 8+27-16\sqrt{2}\]\[-\,\,3\times 9\times 2\sqrt{2}+3\times 3\times 8\] |
| \[=27+72+27+72=198\] |
| Now, \[{{x}^{3}}-\frac{1}{{{x}^{3}}}={{(3+2\sqrt{2})}^{3}}-{{(3-2\sqrt{2})}^{3}}\] |
| \[=27+16\sqrt{2}+3\times 9\times 2\sqrt{2}+3\times 3\times 8-27\] |
| \[+16\sqrt{2}+3\times 9\times 2\sqrt{2}-3\times 3\times 8\] |
| \[=16\sqrt{2}+54\sqrt{2}+16\sqrt{2}+54\sqrt{2}=140\sqrt{2}\] |
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