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JCECE Engineering JCECE Engineering Solved Paper-2003

  • question_answer
    If\[a{{x}^{2}}+bx+c=0\]and\[b{{x}^{2}}+cx+a=0\], then a common root and\[a\ne 0\], then\[\frac{{{a}^{3}}+{{b}^{3}}+{{c}^{3}}}{abc}\]is equal to:

    A) \[1\]                                     

    B) \[2\]

    C)  \[3\]                                    

    D)  \[4\]

    Correct Answer: C

    Solution :

    The condition for common roots gives                 \[{{(bc-{{a}^{2}})}^{2}}={{(ca-b)}^{2}}(ab-{{c}^{2}})\] \[\Rightarrow \]\[{{b}^{2}}{{c}^{2}}+{{a}^{4}}-2{{a}^{2}}bc={{a}^{2}}bc-{{c}^{3}}a-{{b}^{3}}a+{{b}^{2}}{{c}^{2}}\] \[\Rightarrow \]               \[{{a}^{4}}+{{c}^{3}}a+{{b}^{2}}a=3{{a}^{2}}bc\] \[\Rightarrow \]               \[{{a}^{3}}+{{b}^{3}}+{{c}^{3}}=3abc\]


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