Quantitative Aptitude Ques 523
Question: If $ x^{4}+\frac{1}{x^{4}}=727, $ then find the value of $ x^{3}-\frac{1}{x^{3}}. $
Options:
A) 140
B) 120
C) 190
D) 160
Show Answer
Answer:
Correct Answer: A
Solution:
- [a] $ x^{4}+\frac{1}{x^{4}}=727 $ Add 2 on both side, we get, $ x^{4}+\frac{1}{x^{4}}+2=727+2 $
$ \Rightarrow $ $ {{( x^{2}+\frac{1}{x^{2}} )}^{2}}=729 $
$ \Rightarrow $ $ x^{2}+\frac{1}{x^{2}}=\sqrt{729} $
$ \Rightarrow $ $ x^{2}+\frac{1}{x^{2}}+27 $
$ \Rightarrow $ $ {{( x-\frac{1}{x} )}^{2}}+2=27 $
$ \Rightarrow $ $ ,x-\frac{1}{x}=\sqrt{25} $
$ \Rightarrow $ $ x-\frac{1}{x}=5 $
$ \therefore $ $ {{( x-\frac{1}{x} )}^{3}}=x^{3}-\frac{1}{x^{3}}-3,( x-\frac{1}{x} ) $ $ {{(5)}^{3}}=x^{3}-\frac{1}{x^{3}}-3\cdot (5) $
$ \Rightarrow $ $ x^{3}-\frac{1}{x^{3}}=125+15=140 $
BETA
