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Math Help - A good question

  1. #1
    Member great_math's Avatar
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    A good question

    Two numbers are such that the sum of their cubes is 5 and the sum of their squares is 3. Find the sum of the numbers.
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  2. #2
    Moo
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    Hello,
    Quote Originally Posted by great_math View Post
    Two numbers are such that the sum of their cubes is 5 and the sum of their squares is 3. Find the sum of the numbers.
    a^3+b^3=5
    a^2+b^2=3

    Note that a^3+b^3=(a+b)(a^2+b^2-ab)

    \implies 5=(a+b)(3-ab)

    If a and b are integers, then there are a few possibilities :
    3-ab=1 and a+b=5 (1)
    3-ab=5 and a+b=1 (2)
    3-ab=-1 and a+b=-5 (3)
    3-ab=-5 and a+b=-1 (4)

    From (1), we get ab=2, that is a=1 and b=2 for example. But a+b is not equal to 5. Impossible for integers.
    From (2), we get ab=-2. We want a+b=1. --> a=-1 and b=2.
    From (3), we get ab=4 --> a=-1 and b=-4
    From (4), we get ab=8. Impossible for integers.
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Alternatively, let the numbers be x and y. we have

    x^3 + y^3 = 5
    x^2 + y^2 = 3

    So as Moo said, 5 = (x + y)(3 - xy)

    that is, \boxed{5 = 3(x + y) - (x^2y + xy^2)}

    Now, (x + y)^3 = x^3 + y^3 + 3(x^2y + y^2x)

    which means \boxed{5 = (x + y)^3 - 3(x^2y + xy^2)}

    Now, let a = x+ y and b = x^2y + xy^2, then we have the system

    3a - b = 5 .....................(1)
    a^3 - 3b = 5 ....................(2)

    Solve this system for a and you have your answer (since a is the sum of the two numbers)


    I like your solution, Moo

    But, of course, who said they had to be integers?
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  4. #4
    Moo
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    Quote Originally Posted by Jhevon View Post
    Alternatively, let the numbers be x and y. we have

    x^3 + y^3 = 5
    x^2 + y^2 = 3

    So as Moo said, 5 = (x + y)(3 - xy)

    that is, \boxed{5 = 3(x + y) - (x^2y + xy^2)}

    Now, (x + y)^3 = x^3 + y^3 + 3(x^2y + y^2x)

    which means \boxed{5 = (x + y)^3 - 3(x^2y + xy^2)}

    Now, let a = x+ y and b = x^2y + xy^2, then we have the system

    3a - b = 5 .....................(1)
    a^3 - 3b = 5 ................(2)

    Solve this system for a and you have your answer (since a is the sum of the two numbers)


    I like your solution, Moo

    But, of course, who said they had to be integers?
    Yours is far better... I wasn't able to finish off
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  5. #5
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    Thanks for the formulas.
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