11. Solve the equation $27^{x}-7 \sqrt[3]{7 \cdot 3^{x}+6}=6$.

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Solution 1. Let $z=\sqrt[3]{7 y+6}$ and transition to the system

$$
\left\{\begin{array}{l}
y^{3}=7 z+6 \\
z^{3}=7 y+6
\end{array}\right.
$$

Then $y^{3}-z^{3}=7(z-y)$, from which it follows that $y=z$, since $y^{2}+y z+z^{2}$ cannot be equal to -7. As a result, we get the equation $y^{3}-7 y-6=0$, the roots of which are the numbers $-1, -2, 3$. Since $3^{x}>0$, then $3^{x}=3$, hence $x=1$.

Answer: 1.

Topic: "Methods for solving irrational equations"