Make the following statements true by only adding mathematical operators:
\[\begin{align} 0 \quad 0 \quad 0 & = 6 \\ 1 \quad 1 \quad 1 & = 6 \\ 2 \quad 2 \quad 2 & = 6 \\ 3 \quad 3 \quad 3 & = 6 \\ 4 \quad 4 \quad 4 & = 6 \\ 5 \quad 5 \quad 5 & = 6 \\ 6 \quad 6 \quad 6 & = 6 \\ 7 \quad 7 \quad 7 & = 6 \\ 8 \quad 8 \quad 8 & = 6 \\ 9 \quad 9 \quad 9 & = 6 \end{align}\]For example \(2 + 2 + 2 = 6\).
\[\begin{align} (0! + 0! + 0!)! & = 6 \\ (1 + 1 + 1)! & = 6 \\ 2 + 2 + 2 & = 6 \\ 3 \times 3 - 3 & = 6 \\ 4 + 4 - \sqrt{4} & = 6 \\ 5 + 5 / 5 & = 6 \\ 6 + 6 - 6 & = 6 \\ 7 - 7 / 7 & = 6 \\ \left(\sqrt{8 + 8 / 8}\right)! & = 6 \\ (9 + 9) / \sqrt{9} & = 6 \end{align}\]
An interesting thing to note is that if we had four numbers instead of three then for any \(x \ne 0\):
\[\left((x + x + x) / x\right)! = 6\]