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{y+6=(x−3)2x+6=(y−3)2\large{ \begin{cases} y+6 = (x-3)^2 \\ x+6 = (y-3)^2 \end{cases}} ⎩⎨⎧y+6=(x−3)2x+6=(y−3)2
If xxx and yyy are distinct numbers that satisfy the system of equations above, find x2+y2x^2+y^2x2+y2.
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