Multiple Choice Question (MCQ)

If a+b+c=0 and a^2+b^2+c^2=1, then a^4+b^4+c^4 is

  1. ×

    \dfrac{1}{8}

  2. ×

    \dfrac{1}{4}

  3. \dfrac{1}{2}

  4. ×

    \dfrac{1}{3}

Collected in the board: a+b+c=0 problems

Steven Zheng posted 1 month ago


Answer

  1. \because a +b+c=0

    Using the square of a trinomial formula

    (a+b+c)^2 = a^2+b^2+c^2+2(ab+bc+ac)

    \therefore a^2+b^2+c^2+2(ab+bc+ac) = 0


    \because a^2+b^2+c^2=1

    \therefore ab+bc+ac = -\dfrac{1}{2}

    Square both sides of the equation

    a^2b^2+b^2c^2+a^2c^2+2abc(a+b+c) = \dfrac{1}{4}

    Therefore,

    a^2b^2+b^2c^2+a^2c^2 = \dfrac{1}{4}

    Then,

    a^4+b^4+c^4 = ( a^2+b^2+c^2)^2 - 2( a^2b^2+b^2c^2+a^2c^2)

    = 1 - 2\times \dfrac{1}{4} = \dfrac{1}{2}

    C is the choice


Steven Zheng posted 1 month ago

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