Railways Quantitative Aptitude Power, Surds And Indices Sample Paper Surds Indices Sample Test Paper-4

  • question_answer
    If \[a={{x}^{\frac{1}{3}}}+{{x}^{\frac{1}{3}}}\]  then \[{{a}^{3}}-3a=\]

    A)  \[x-{{x}^{-1}}\]         

    B)  \[2x\]

    C)  \[x+{{x}^{-1}}\]       

    D)  0

    Correct Answer: C

    Solution :

    [c] \[a={{x}^{\frac{1}{3}}}+{{x}^{\frac{1}{3}}}\] Cubing both sides, we get \[{{a}^{3}}=x+\frac{1}{x}+3({{x}^{\frac{1}{3}}}+{{x}^{\frac{1}{3}}})\] \[{{a}^{3}}=x+\frac{1}{x}+3a\] \[{{a}^{3}}-3a=x+{{x}^{-1}}\]


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