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{2x3=1+y+1232y3=1+x+123\begin{cases} 2x^3=1+\sqrt[3]{\dfrac{y+1}{2}} \\ 2y^3=1+\sqrt[3]{\dfrac{x+1}{2}}\end{cases}⎩⎪⎪⎨⎪⎪⎧2x3=1+32y+12y3=1+32x+1 Given the system of equations above, find x+yx+yx+y.
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