SSC Quantitative Aptitude Algebra Question Bank Elementary Algebra (I)

  • question_answer
    \[\left( x+\frac{1}{x} \right)\left( x-\frac{1}{x} \right)\left( {{x}^{2}}+\frac{1}{{{x}^{2}}}-1 \right)\left( {{x}^{2}}+\frac{1}{{{x}^{2}}}+1 \right)\]is equal to

    A) \[{{x}^{6}}+\frac{1}{{{x}^{6}}}\]

    B) \[{{x}^{8}}+\frac{1}{{{x}^{8}}}\]

    C) \[{{x}^{8}}-\frac{1}{{{x}^{8}}}\]

    D) \[{{x}^{6}}-\frac{1}{{{x}^{6}}}\]

    Correct Answer: D

    Solution :

    [d] \[\left( x+\frac{1}{x} \right)\left( x-\frac{1}{x} \right)\] \[\left( {{x}^{2}}+\frac{1}{{{x}^{2}}}1- \right)\left( {{x}^{2}}+\frac{1}{{{x}^{2}}}+1 \right)\] \[=\left( {{x}^{2}}-\frac{1}{{{x}^{2}}} \right)\left[ {{\left( {{x}^{2}}+\frac{1}{{{x}^{2}}} \right)}^{2}}-1 \right]\] \[=\left( {{x}^{2}}-\frac{1}{{{x}^{2}}} \right)\left( {{x}^{4}}+\frac{1}{{{x}^{4}}}+1 \right)\] \[={{x}^{6}}-\frac{1}{{{x}^{6}}}\]


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