Why be a square when you can be a cube?

Solve the following equation over integral values of \(y,x\):

\(y^3=x^3+8x^2-6x+8\)

Details and assumptions:

  • Please input your answer as the sum of all possible values for \((x,y)\).As an explicit example,if the ordered pair \((1,2)\) is an answer and is the only one,input \(3\) but if \((2,3)\) is also an answer, input \( 8 \)

  • \(x,y \in \mathbb {Z}^+\)

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