jmc

We hope that we can write $52+6\sqrt{43}$ as the square of some expression of the form $a+b\sqrt{43},$ where $a$ and $b$ are integers. To find $a$ and $b,$ we write $$52+6\sqrt{43}=(a+b\sqrt{43}{)}^2=(a^2+43b^2)+2ab\sqrt{43}.$$Therefore, $a^2+43b^2=52$ and $2ab=6.$ Testing pairs $(a,b)$ such that $2ab=6,$ we find a solution, $(a,b)=(3,1),$ so indeed, $$\sqrt{52+6\sqrt{43}}=3+\sqrt{43}.$$Similarly, we have $$\sqrt{52-6\sqrt{43}}=-3+\sqrt{43}$$(remembering to take the positive square root). Now we can compute the answer: $$\begin{array}{rl}(52+6\sqrt{43}{)}^{3/2}-(52-6\sqrt{43}{)}^{3/2}& =(52+6\sqrt{43})(3+\sqrt{43})-(52-6\sqrt{43})(-3+\sqrt{43})\\ & =2\cdot 52\cdot 3+2\cdot 6\cdot 43\\ & =\boxed{828}.\end{array}$$