# 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 is $$(2,3)$$ also an answer input $$8$$

• $$x,y \in \mathbb {Z}^+$$

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