Reaction between alumina and aqueous sodium hydroxide, which one's right? My chemistry textbook has a rather annoying irregularity.

It cites two different reactions for the same set of reactants (aluminium oxide, sodium hydroxide and water) without the slightest mention of a particular temperature or concentration. Here they are:


In a chapter on the p-block elements:


  $$\ce{Al2O3 (s) + 6 NaOH (aq) + 3 H2O (l) -&gt; 2 Na3[Al(OH)6] (aq)}$$

In a chapter on metallurgy, under the topic of 'Extraction of aluminium from bauxite':


  $$\ce{Al2O3 (s) + 2 NaOH (aq) + 3 H2O (l) -&gt; 2 Na[Al(OH)4] (aq)}$$



Which one of these is correct? Or are they both correct, depending on the situation? If so, under what conditions are they correct?
Quoting from chemguide (emphasis mine):


  With hot concentrated sodium hydroxide solution, aluminium oxide reacts to give a solution of sodium tetrahydroxoaluminate.
  
  $$\ce{Al2O3(s) + 2NaOH(aq) + 3H2O(l) -&gt; 2NaAl(OH)4}$$
  
  Note: You may find all sorts of other formulae given for the product from this reaction. These range from $\ce{NaAlO2}$ (which is a dehydrated form of the one in the equation) to $\ce{Na3Al(OH)6}$ (which is a different product altogether). 
  
  What you actually get will depend on things like the temperature and the concentration of the sodium hydroxide solution. In any case, the truth is almost certainly a lot more complicated than any of these.
  
  The version I am using is perfectly acceptable and is consistent with the aluminium chemistry you will find elsewhere on the site.


You can see that products of the reaction depends on the concentration of sodium hydroxide (Basic pH) but  other factors like temperature and pressure also come into play.

If you do not add water and only change the concentration of sodium hydroxide, product will also be different:

$$
\begin{align}
\ce{Al2O3 + 6 NaOH &amp;-&gt; 2 Al(OH)3 + 3 Na2O}\\
\ce{Al2O3 + 2NaOH &amp;-&gt;[\pu{900 .. 1100 ^\circ C}] 2 NaAlO2 + H2O}
\end{align}
$$

Sodium aluminate is generally $\ce{NaAlO2}$ but its aqueous solution contains many other species like $\ce{[Al(OH)4]−}$, $\ce{[Al(OH)6]^3-}$ or perhaps the $\ce{[Al(H2O)2(OH)4]−}$. They are all considered to be aluminate species. So, if you are getting sodium tetrahydroxoaluminate or sodium hexahydroxoaluminate, they are no different compounds but are all hydrated form of sodium aluminate (addition of water + change in the concentration of sodium hydroxide).