8th Class Mathematics Factorisation Question Bank Factorization

  • question_answer
    The value of \[\left( x-y \right)\left( x+y \right)\left( {{x}^{2}}+{{y}^{^{2}}} \right)\left\{ {{\left( {{x}^{2}}+{{y}^{^{2}}} \right)}^{2}}-2{{x}^{2}}{{y}^{2}} \right\}\]is

    A)  \[{{x}^{8}}-{{y}^{8}}\]      

    B)  \[{{x}^{8}}+{{y}^{8}}\]

    C)  \[{{x}^{6}}-{{y}^{6}}\]                   

    D)  \[{{x}^{6}}+{{y}^{6}}\]

    Correct Answer: A

    Solution :

    (a): \[\left( x-y \right)\left( x+y \right)\left( {{x}^{2}}+{{y}^{2}} \right)\left\{ \left( {{x}^{2}}+{{y}^{2}} \right)-2{{x}^{2}}{{y}^{2}} \right\}\]Combining from left in bunches of two factors each, \[\left( x-y \right)\left( x+y \right)={{x}^{2}}-{{y}^{2}}\] Then\[\left( {{x}^{2}}-{{y}^{2}} \right)\left( {{x}^{2}}+{{y}^{2}} \right)={{x}^{4}}-{{y}^{4}}\] Then, \[\left( {{x}^{4}}-{{y}^{4}} \right)\left( {{x}^{4}}+{{y}^{4}} \right)={{x}^{8}}-{{y}^{8}}\] \[\left[ since\left\{ {{({{x}^{2}}+{{y}^{2}})}^{2}}-2{{x}^{2}}{{y}^{2}} \right\}={{x}^{4}}+{{y}^{4}} \right]\].


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