Solution: The necessary and sufficient condition for a quartic equation to be reducible to a quadratic equation, as established in problem 656 (October 1955 issue, p. 57), is satisfied in this case. Indeed,

$$a=\frac{18}{27}=\frac{2}{3}, \quad b=0, \quad c=-\frac{1}{27}$$

and thus

$$a^{3}-4 a b+8 c=\frac{8}{27}-\frac{8}{27}=0$$

Using the transformation $x=z-\frac{a}{4}=z-\frac{1}{6}$ given there, we obtain the equation

$$\left(z-\frac{1}{6}\right)^{4}+\frac{2}{3}\left(z-\frac{1}{6}\right)^{3}-\frac{1}{27}\left(z-\frac{1}{6}\right)+1=z^{4}-\frac{1}{6} z^{2}+\frac{1301}{1296}=0$$

The discriminant of this equation, which is quadratic in $z^{2}$, is negative, and thus it cannot have real roots.

Second solution: Notice that if we multiply our equation by 12 and add and subtract $36 x^{2}$ to the left side, the left side can be completed to a perfect square:

$$324 x^{4}+216 x^{3}+36 x^{2}-36 x^{2}-12 x+1=-11$$

or

$$\left(18 x^{2}+6 x-1\right)^{2}=-11$$

The left side cannot be negative for real $x$, while the right side is negative, so our equation cannot be satisfied by any real root.