Quantitative Aptitude Ques 1374
Question: $ x=3+2\sqrt{2}, $ then the values of $ x^{3}+\frac{1}{x^{3}} $ and $ x^{3}-\frac{1}{x^{3}} $ are respectively
Options:
A) $ 140\sqrt{2}, $ $ 198 $
B) $ 234, $ $ 216 $
C) $ 216, $ $ 234 $
D) $ 198, $ $ 140\sqrt{2} $
Show Answer
Answer:
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} $
BETA
