Let $s$ be an integer greater than $6$. A solid cube of side $s$ has a square
hole of side $x < 6$ drilled directly through from one face to the opposite
face (so the drill removes a cuboid). The volume of the remaining solid
is numerically equal to the total surface area of the remaining solid.
Determine all possible integer values of $x$.

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I've figured out the first bit, setting the equations. Would it be easier if I said "Find the solutions of $s$ and $x$ such that $s^3-x^2s=6s^2-2x^2+4xs$"? (bearing in mind that $s$ must be $>6$ and an integer and $x$ must be $<6$).