Find all values of $x$ such that
\[
\begin{vmatrix} 1 & 1 & 1 \\ x & 7 & -2 \\ x^3 & 343 & -8 \end{vmatrix}
= 0.\]Enter your values separated by commas.  For example, if you think the possible values are 4, 5, and 6, then enter you answer as "4,5,6".
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 $-7 - (-2) = -5.$)