11. Given that $a$ and $b$ are real numbers, satisfying:
$$
\sqrt[3]{a}-\sqrt[3]{b}=12, \quad a b=\left(\frac{a+b+8}{6}\right)^{3} \text {. }
$$

Then $a-b=$ $\qquad$ (Proposed by Thailand)
11. 468 .

Let $x=\sqrt[3]{a}, y=\sqrt[3]{b}$. Then $x-y=12, 6xy=x^{3}+y^{3}+8$.
Thus, $x^{3}+y^{3}+2^{3}-3 \times 2xy=0$
$$
\Rightarrow(x+y+2)\left(x^{2}+y^{2}+2^{2}-xy-2x-2y\right)=0 \text {. }
$$

Therefore, $x+y+2=0$, or
$$
x^{2}-xy+y^{2}-2x-2y+4=0 \text {. }
$$

Equation (1) can be rewritten as
$$
\frac{1}{2}(x-y)^{2}+\frac{1}{2}(x-2)^{2}+\frac{1}{2}(y-2)^{2}=0 \text {. }
$$

Thus, $x=y=2$, but this does not satisfy $x-y=12$.
Hence, $x+y+2=0$.
Therefore, $x=5, y=-7$.
Thus, $a=125, b=-343, a-b=468$.