Solve the following equation:

$$
\frac{1+x^{4}}{(1+x)^{4}}=\frac{3}{4}
$$
Solution. The denominator cannot be zero, so $x \neq -1$. After multiplying by the denominators and expanding the fourth power of the binomial polynomial,

$$
4+4 x^{4}=3\left(1+4 x+6 x^{2}+4 x^{3}+x^{4}\right)
$$

further multiplication and rearrangement to zero yield

$$
x^{4}-12 x^{3}-18 x^{2}-12 x+1=0
$$

This is a symmetric quartic equation, often referred to as a "reciprocal equation," which can always be reduced to a quadratic equation. The equation clearly does not have $x=0$ as a root, so we can divide by $x^{2}$:

$$
x^{2}-12 x-18-12 \frac{1}{x}+\frac{1}{x^{2}}=0
$$

Now, introduce a new variable for the expression $x+\frac{1}{x}$. With the substitution $x+\frac{1}{x}=a$, and squaring and rearranging,

$$
\left(x+\frac{1}{x}\right)^{2}=a^{2} \Rightarrow x^{2}+\frac{1}{x^{2}}=a^{2}-2
$$

Transform the equation using the new variable:

$$
\begin{aligned}
x^{2}+\frac{1}{x^{2}}-12\left(x+\frac{1}{x}\right)-18 & =0 \\
a^{2}-12 a-20 & =0
\end{aligned}
$$

The roots of the equation are $a_{1,2}=6 \pm 2 \sqrt{14}$. The absolute value of $a=6-2 \sqrt{14}$ is less than 2, so it cannot be the sum of a real number and its reciprocal. Real solutions can only be obtained for $a=6+2 \sqrt{14}$.

$$
\begin{aligned}
x+\frac{1}{x} & =6+2 \sqrt{14} \Rightarrow x^{2}-(6+2 \sqrt{14}) x+1=0 \\
x_{1,2} & =3+\sqrt{14} \pm \sqrt{22+6 \sqrt{14}}
\end{aligned}
$$

The solutions indeed satisfy the equation.

Note. The problem statement did not specify the domain, so those who determined the complex roots also received full marks. These roots come from the value $a=6-2 \sqrt{14}$:

$$
x_{3,4}=3-\sqrt{14} \pm i \sqrt{6 \sqrt{14}-22}
$$