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KVPY Sample Paper KVPY Stream-SX Model Paper-15

  • question_answer
     \[x+y+z=12\] &  \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}=96\] & \[\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=36\] then value of \[{{x}^{3}}+{{y}^{3}}+{{z}^{3}}\] is

    A) 862

    B) 863  

    C) 868

    D) 866

    Correct Answer: D

    Solution :

    \[{{(x+y+z)}^{2}}=144\]
    \[\therefore \sum{{{x}^{2}}+2\sum{x}y=144}\]
    \[\Rightarrow \sum{xy=24}\]
    \[\frac{\sum{xy}}{xyz}=36\]
    \[\Rightarrow xyz=\frac{2}{3}\]
    \[\therefore {{x}^{3}}+{{y}^{3}}+{{z}^{3}}-3xyz=(x+y+z)\,\,(\sum{{{x}^{2}}-\sum{xy)}}\]
    \[\therefore \sum{{{x}^{3}}-2=12\,\,(96-24)}\]
    \[\Rightarrow \sum{{{x}^{3}}=866}\]


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