Quantitative Aptitude Ques 1692
Question: If $ ( a+\frac{1}{a} )=4\sqrt{2}, $ then what is the value of $ (a^{6}+{a^{-6}}) $ ?
Options:
A) 26910
B) 25800
C) 2400
D) 1390
Show Answer
Answer:
Correct Answer: A
Solution:
- Given, $ ( a+\frac{1}{a} )=4\sqrt{2} $ On squaring both sides, we get $ {{( a+\frac{1}{a} )}^{2}}={{(4\sqrt{2})}^{2}} $
$ \Rightarrow $ $ {{(4\sqrt{2})}^{2}}=a^{2}+\frac{1}{a^{2}}+2 $ $ [\because {{(a+b)}^{2}}=a^{2}+b^{2}+2ab] $
$ \Rightarrow $ $ a^{2}+\frac{1}{a^{2}}=32-2=30 $ Now, on cubing both sides, we get $ {{( a^{2}+\frac{1}{a^{2}} )}^{3}}={{(30)}^{3}} $
$ \Rightarrow $ $ a^{6}+\frac{1}{a^{6}}+3\cdot a^{2}\times \frac{1}{a^{2}}( a^{2}+\frac{1}{a^{2}} )=27000 $ $ [\because {{(a+b)}^{3}}=a^{3}+b^{3}+3ab,(a+b)] $
$ \Rightarrow $ $ a^{6}+\frac{1}{a^{6}}+3(30)=27000 $
$ \Rightarrow $ $ a^{6}+\frac{1}{a^{6}}=27000-90 $
$ \Rightarrow $ $ a^{6}+\frac{1}{a^{6}}=26910 $
$ \therefore $ $ a^{6}+{a^{-6}}=26910 $
BETA
