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SSC Sample Paper Mock Test-16 SSC CGL Tear-II Paper-1

  • question_answer
       If\[{{x}^{2}}-3x+1=0,\]then the value of \[{{x}^{3}}+\frac{1}{{{x}^{3}}}\]

    A) 9

    B)  18     

    C)  27

    D)  1       

    Correct Answer: B

    Solution :

         \[{{x}^{2}}-3x+1=0\] \[\Rightarrow \]   \[{{x}^{2}}+1=3x\]\[\Rightarrow \]\[x+\frac{1}{x}=3\] \[\therefore \]      \[{{x}^{3}}+\frac{1}{{{x}^{3}}}={{\left( x+\frac{1}{x} \right)}^{3}}-3x\cdot \frac{1}{x}\left( x+\frac{1}{x} \right)\] \[=27-3\times 3=18\] 


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