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Math Help - a^4+b^4+c^4

  1. #1
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    a^4+b^4+c^4

    Suppose a,b,c satisfy the following equation:
    a+b+c=3;
    a2+ b2 +c2= 5;
    a3+b3+c3=7;
    what is a4+b4+c4







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  2. #2
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    Quote Originally Posted by kamaksh_ice View Post
    Suppose a,b,c satisfy the following equation:
    a+b+c=3;
    a2+ b2 +c2= 5;
    a3+b3+c3=7;
    what is a4+b4+c4

    SEND ME DE ENTIRE SOLUTION
    Hello,


    I assume that you mean aČ when you write a2.

    I can't give you the complete solution, only some considerations. But maybe you can use them for a start:

    (a+1)^4+(b+1)^4+(c+1)^4 = a^4+4a^3+6a^2+4a+1 + b^4+4b^3+6b^2+4b+1+c^4+4c^3+6c^2+4c+1

    a^4+b^4+c^4+4(a^3+b^3+c^3)+6(a^2+b^2+c^2)+4(a+b+c)  +3 According to the problem this term can be reduced to:

    a^4+b^4+c^4+4 \cdot (7)+6 \cdot (5)+4 \cdot (3)+3 = a^4+b^4+c^4+73

    And now I need a small spark of inspiration.
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  3. #3
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    a^2+b^2+c^2 = 5
    Square,
    a^4+b^4+c^4 + 2(a^2b^2+b^2c^2+a^2c^2)=25...(1)


    a+b+c=3
    Square,
    a^2+b^2+c^2+2(ab+ac+bc)=9
    Thus,
    ab+bc+ac=2
    Square,
    a^2b^2+b^2c^2+a^2c^2+2(a^2bc+ab^2c+abc^2)=4
    Thus,
    a^2b^2+b^2c^2+a^2c^2+2abc(a+b+c)=4
    Thus,
    a^2b^2+b^2c^2+a^2c^2+6abc = 4...(2)

    Use the identity,
    a^3+b^3+c^3 - 3abc = (a^2+b^2+c^2 - ab - bc - ac)(a+b+c)
    Thus,
    7 - 3abc = (5-2)(3)
    Thus,
    6abc=-4...(3)

    Thus by (2) and (3) we have,
    a^2b^2+b^2c^2+a^2c^2=8

    Thus by (1) we have,
    a^4+b^4+c^4+2(8)=25
    Thus,
    a^4+b^4+c^4=9
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by kamaksh_ice View Post
    Suppose a,b,c satisfy the following equation:
    a+b+c=3;
    a2+ b2 +c2= 5;
    a3+b3+c3=7;
    what is a4+b4+c4
    You may, of course, solve the system by substitution. Unfortunately when I do this I can't get an exact answer for a without using Cardano's method.

    -Dan
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