Quantitative Aptitude Ques 1240
Question: If $ x+\frac{1}{x}=\sqrt{3}, $ then the value of $ x^{18}+x^{12}+x^{6}+1 $ is
Options:
A) 0
B) 1
C) 2
D) 8
Show Answer
Answer:
Correct Answer: A
Solution:
- Given, $ x+\frac{1}{x}=\sqrt{3} $ (i) Cubing on both sides, we get $ {{( x+\frac{1}{x} )}^{3}}={{(\sqrt{3})}^{3}} $ $ [\because {{(a+b)}^{3}}=a^{3}+b^{3}+3ab(a+b)] $
$ \Rightarrow $ $ x^{3}+\frac{1}{x^{3}}+3( x+\frac{1}{x} )={{(\sqrt{3})}^{3}} $ [from Eq.(i)]
$ \Rightarrow $ $ x^{3}+\frac{1}{x^{3}}+3\sqrt{3}=3\sqrt{3} $
$ \Rightarrow $ $ x^{3}+\frac{1}{x^{3}}=0 $
$ \Rightarrow $ $ x^{16}+x^{12}+x^{6}+1=x^{12}(x^{6}+1)+1(x^{2}+1) $ $ =(x^{12}+1)(x^{6}+1) $ $ =(x^{12}+1)\cdot x^{3}( x^{3}+\frac{1}{x^{3}} )=0 $
BETA
