Question 903510
x and y are the original rectangle dimensions; the area for the bottom after cutting the a inch squares is (x-2a)(y-2a).  The volume of the box will be 
{{{(x-2a)(y-2a)a=528}}};


"Length is twice its width" for rectangle's dimensions:
If x is length, and y is width, then x=2y, and {{{(2y-2a)(y-2a)a=528}}}
{{{2y^2-2ay-4ay+4a^2-528=0}}}
{{{2y^2-6ay+4a^2-528=0}}}
{{{highlight(y^2-3ay+2a^2-264=0)}}}
Continue solving for y using the general solution to a quadratic formula, and use the y result to get x=2y.  That will be for any "a" in general.  


Once that is done, you can substitute the a=3 and evaluate that more specific case.