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    Mark Posts: 1, Reputation: 1
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    #1

    Mar 31, 2003, 10:05 PM
    Algebra
    Can you please help me solve?
    If M+N=3 and M^2+N^2=6, Find the numerical value for
    M(cubed)+N (cubed).
    MathMaven53's Avatar
    MathMaven53 Posts: 20, Reputation: 2
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    #2

    Apr 1, 2003, 04:15 PM
    M^2 + N^2 M^2 + N^2 Re: Algebra
    We are given M+N=3 and M^2 + N^2 =6

    Factor M^3 + N^3

    M^3 + N^3 = (M+N)(M^2 -MN + N^2)

    = (M+N)(M^2 + N^2 - MN)

    We have all quantities on the right side except MN

    To find it, square M + N = 3

    M^2 + 2MN + N^2 = 9

    Substituting for M^2 + N^2 and solving for MN gives

    MN= 3/2

    Then


    M^3 + N^3 = (M+N)(M^2 + N^2 - MN)

    = 3(6 - 3/2)

    = 27/2

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