Fermat's Cousin

The Fermat point minimizes the sum of its distance from the vertices of a triangle. The centroid minimizes the sum of the squares of its distance from the vertices. In $\triangle ABC$ , can you find the point which minimizes the sum of the $cube$ of its distance from the vertices? That is, find the point $Q$ that minimizes $AQ^3 +BQ^3 + CQ^3$ and submit its distance from $A$ . A closed-form for $AQ$ exists. Find it and submit $\lfloor{10^6\cdot{AQ}}\rfloor$ .