Subtracting the second column from the first column, we get \[ \begin{vmatrix} 0 & 1 & 1 \\ x - 7 & 7 & -2 \\ x^3 - 343 & 343 & -8 \end{vmatrix} = 0.\]The first column becomes all zeros when $ x = 7, $ so this is one possible value of $ x $. Subtracting the third column from the first column, we get \[ \begin{vmatrix} 0 & 1 & 1 \\ x + 2 & 7 & -2 \\ x^3 + 8 & 343 & -8 \end{vmatrix} = 0.\]The first column becomes all zeros when $ x = -2, $ so this is another possible value of $ x $. Expanding the last determinant along the first column, we get \begin{align*} \begin{vmatrix} 0 & 1 & 1 \\ x + 2 & 7 & -2 \\ x^3 + 8 & 343 & -8 \end{vmatrix} &= -(x + 2) \begin{vmatrix} 1 & 1 \\ 343 & -8 \end{vmatrix} + (x^3 + 8) \begin{vmatrix} 1 & 1 \\ 7 & -2 \end{vmatrix} \\ &= -9x^3 + 351x + 630 = 0.\end{align*}Since we know that $ x = 7 $ are $ x = -2 $ are two solutions, we can take out a factor of $ x - 7 $ and $ x + 2 $, to get \[-9(x - 7)(x + 2)(x + 5) = 0.\]Hence, the possible values of $ x $ are $ \boxed{7, -2, \text{ and } -5} $. (Alternatively, by Vieta's formulas, the sum of the three roots of $ -9x^3 + 351x + 630 = 0 $ is 0, so the third root is $ -7 - (-2) = -5 $.)