Answer: 828

Small Hint:

Express $52\pm6\sqrt{43}$ as squares of conjugate radical expressions

Big Hint:

After taking the $\frac{3}{2}$ powers, expand the difference of the two cubes symmetrically

Solution:

Since $$52\pm6\sqrt{43}=(\sqrt{43}\pm3)^2$$ and $\sqrt{43}\gt3,$ the expression is $(\sqrt{43}+3)^3-(\sqrt{43}-3)^3.$ Using $(x+y)^3-(x-y)^3=6x^2y+2y^3$ with $x=\sqrt{43}$ and $y=3$ gives $$\begin{aligned}6(43)(3)+2(27)&=774+54\\&=828.\end{aligned}$$