inorganic chemistry - Reaction of aluminium trichloride with water

Since $\ce{Al^3+}$ is an acidic cation, shouldn't it react with water like this? \begin{align} \ce{AlCl3 + 6H2O &<=> [Al(H2O)6]^3+ + 3Cl^-} \tag1\\ \ce{[Al(H2O)6]^3+ +H2O &<=> [Al(H2O)5(OH)]^2+ + H3O^+} \tag2\\ \ce{[Al(H2O)5(OH)]^2+ + H2O &<=> [Al(H2O)4(OH)2]^+ + H3O^+} \tag3\\ \ce{[Al(H2O)4(OH)2]^+ +H2O &<=> [Al(H2O)3(OH)3](s) + H3O^+} \tag4\\ \end{align}

The 3rd step of the equilibrium rarely happens in water. But if we add a base, strong or not ($\ce{NaOH}$, or $\ce{Na2S}$) the equilibrium weighs to the right and $\ce{Al(OH)3}$ is formed.

But I've seen people write it as this. $$ \ce{AlCl3 + 3H2O -> Al(OH)3 + 3HCl}$$ Is this correct? I need a certain answer.