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NTSE Sample Paper NTSE SAT Practice Test-19

  • question_answer
    If \[x=(b-c)(a-d);y=(c-a)\]\[(b-d);z=(a-b)(c-d)\] then the value of\[{{x}^{3}}+{{y}^{3}}+{{z}^{3}}\]is

    A) \[\frac{x\,y\,z}{3}\]                   

    B) \[2x\,y\,z\]

    C) \[3x\,y\,z\]         

    D) \[-3x\,y\,z\]

    Correct Answer: C

    Solution :

    [c] It can be seen that \[x+y+z=0\] \[\therefore \,\,\,{{x}^{3}}+{{y}^{3\text{ }}}+\text{ }{{z}^{3}}=3\text{ }x\text{ y }z.\]        


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