Question 1209170
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Find one ordered pair (x,y) of real numbers such that x + y = 8 and x^3 + y^3 = 200 - 4x^2 - 4y^2.
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Left side  x^3 + y^3  can be composed  as  (x+y)*(x^2 - xy + y^2).

The factor  (x+y)  can be replaced by 8,  based on the first equation.


After that, second equation can be written in this form

    8(x^2 - xy + y^2) = 200 - 4(x^2 + y^2),

or

    12(x^2 + y^2) - 8xy = 200,

    12(x+y)^2 - 24xy - 8xy = 200,

    12(x+y)^2 - 32xy = 200.


Again, we can replace here  (x+y)  by 8, and we get then

    12*8^2 - 32xy = 200,

    32xy = 12*64 - 200 = 568,

      xy = 568/32 = 17.75.


But under the condition x+y = 8,  well known minimax property says 
that the product xy may have the maximum value of  4*4 = 16  at  x = y = 8/2 = 4.


In other words, under given conditions, the original system has no real solutions.


So, there is no any pair of real numbers (x,y) satisfying the imposed conditions.
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Solved completely in Math terms.