.
Find one ordered pair (x,y) of real numbers such that x + y = 8 and x^3 + y^3 = 200 - 4x^2 - 4y^2.
~~~~~~~~~~~~~~~~~~~~~~~
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.
Solved completely in Math terms.