I found the Smith normal form of a matrix $A$ to be
$$B=\begin{pmatrix} 1 & 0 &0\\ 0 & 4 & 0\\ 0&0& 8 \end{pmatrix}$$
Can I conclude now that $\mathbb{Z}^3\big/ A\mathbb{Z}^3=\mathbb{Z}^3\big/ B\mathbb{Z}^3$?
And is it right to calculate $\mathbb{Z}^3\big/ B\mathbb{Z}^3=(\mathbb{Z}\times\mathbb{Z}\times\mathbb{Z})\big/ (\mathbb{Z}\times4\mathbb{Z}\times8\mathbb{Z})=\{0\}\times\mathbb{Z}/4\mathbb{Z}\times\mathbb{Z}/8\mathbb{Z}$?