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JEE Main & Advanced Mathematics Complex Numbers and Quadratic Equations Question Bank De Moivre's theorem and Roots of unity

  • question_answer
    If \[1,\omega ,{{\omega }^{2}}\] are three cube roots of unity, then \[{{(a+b\omega +c{{\omega }^{2}})}^{3}}\] + \[{{(a+b{{\omega }^{2}}+c\omega )}^{3}}\] is equal to, if \[a+b+c=0\] [West Bengal JEE 1992]

    A) \[27\,abc\]

    B) 0

    C) \[3\,abc\]

    D) None of these

    Correct Answer: A

    Solution :

    Trick: Put\[a=1,b=1,c=-2\], \[\because \]  \[a+b+c=0\] \[\therefore \,\,\,{{(1+\omega -2{{\omega }^{2}})}^{3}}+{{(1+{{\omega }^{2}}-2\omega )}^{3}}\] \[={{(-3{{\omega }^{2}})}^{3}}+{{(-3\omega )}^{3}}=-27-27=-54\] Also option (a) gives the value \[i.e.,\]\[27\times 1\times 1(-2)=-54\]


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