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Maths factorization problem

  1. Mar 6, 2006 #1
    Hello,
    I have got one mathematical problem and i am not able to solve it may i plese get some help. My sum is as foolws:-
    a^2+b^2=25 and a^3+b^3=91 so now find the values of a and b.
    Now here we cannot tahe into consideration that as a^2+b^2=25 , a=3 and b=4 or a=4 and b=3 .But we have to find it by mathematical formulas and without any assumptions. thank you in advance. Please send it soon
     
  2. jcsd
  3. Mar 7, 2006 #2
    [tex] a^2 + b^2 = 25 \therefore a = \sqrt{25 - b^2} [/tex]
    [tex] a = (91 - b^3)^{1/3} [/tex]
    Combine the two and solve to get a value (possibly more than one value), and then use these to work out value(s) for a. The re-arranging might be a little complex, but it should be do-able.
     
    Last edited by a moderator: Mar 7, 2006
  4. Mar 7, 2006 #3

    VietDao29

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    Homework Helper

    [tex]\left\{ \begin{array}{l} a ^ 2 + b ^ 2 = 25 \quad (1) \\ a ^ 3 + b ^ 3 = 91 \quad (2) \end{array} \right.[/tex]
    I think you may want to try this:
    Now let x = a + b, and y = ab, we will try to write the equation (1), and (2) in terms of x, and y:
    a2 + b2 = (a + b)2 - 2ab = x2 - 2y
    a3 + b3 = (a + b) (a2 - ab + b2) = (a + b) ((a + b)2 - 3ab) = x (x2 - 3y) = x3 - 3xy.
    So you'll have:
    [tex]\left\{ \begin{array}{l} x ^ 2 - 2y = 25 \quad (3) \\ x ^ 3 - 3xy = 91 \quad (4) \end{array} \right.[/tex]
    Now from the equation (3), one can solve y in terms of x, then plug y in equation (4), and solve for x. From there, you can solve for y.
    having x = a + b, and y = ab, one then can find a, and b.
    Can you go from here? :)
     
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