Quantitative Aptitude Ques 581

Question: $ x^{2}+4x+3=0, $ then the value of $ \frac{x^{3}}{x^{6}+27x^{3}+27} $ is

Options:

A) $ -1 $

B) $ -\frac{1}{2} $

C) 1

D) $ \frac{1}{2} $

Show Answer

Answer:

Correct Answer: A

Solution:

  • $ x^{2}+4x+3=0 $

$ \Rightarrow $ $ x^{2}+3x+x+3=0 $

$ \Rightarrow $ $ x,(x+3)+1,(x+3)=0 $

$ \Rightarrow $ $ x=-,3 $ and $ -1 $ Now, $ \frac{x^{3}}{x^{6}+27x^{3}+27}=\frac{x^{3}}{x^{3}( x^{3}+27+\frac{27}{x^{3}} )} $ $ =\frac{1}{x^{3}+27+\frac{27}{x^{3}}} $ On putting $ x=-,3 $ we get $ \frac{1}{{{(-,3)}^{3}}+27+\frac{27}{{{(-,3)}^{3}}}}=-1 $

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