Lesson A truly miraculous evaluating problem with a truly miraculous solution

A truly miraculous evaluating problem with a truly miraculous solution

Problem 1

We are given this equation

    (x^2+1)*(y^2+1) + 25 = 10*(x+y).    (1)


Let's transform it equivalently step by step to something more expressive.

Open parentheses

    x^2*y^2 + x^2 + y^2 + 1 = 10(x+y) - 25,

    x^2*y^2 + (x^2 + 2xy + y^2) - 2xy + 1 = 10*(x+y) - 25, 

    x^2*y^2 + (x+y)^2 - 2xy + 1 = 10*(x+y) - 25,

    (x^2*y^2 - 2xy + 1) + (x+y)^2 = 10*(x+y) - 25

    (xy-1)^2 = -(x+y)^2 + 10*(x+y) - 25,

    (xy-1)^2 = -((x+y)^2 - 10*(x+y) + 25),

    (xy-1)^2 = -((x+y)-5)^2.


    +----------------------------------------------------------+
    |   On the left side, we have a square of a real number.   |
    |   On the right side, we have a square of a real number   |
    |            with the NEGATIVE sign.                       |
    +----------------------------------------------------------+


       Such an equality can be true IF and ONLY IF the numbers 
    in the left side and in the right side both are equal to zero.


So, we deduced from equation (1), that

    xy = 1,     (2)

    x+y = 5.    (3)


Again, a miracle had happened: from one complicated equation (1) for two unknowns x and y, 
we deduced two simple equations (2) and (3) for these unknowns.


         This miracle happened due to the special form of equation (1).


Now we are in position to calculate  x^3 + y^3.


    x^3 + y^3 = (x+y)*(x^2 - xy + y^2 ) = (x+y)*((x^2 + 2xy + y^2) - 3xy) = (x+y)*((x+y)^2 - 3xy) = 

          Now substitute here xy = 1  and  x+y = 5  from (2) and (3), and continue

    = 5*(5^2 - 3*1) = 5*(25-3) = 5*22 = 110.


Therefore,   =  = .


ANSWER.   = .

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