## Task B-1.2.

For which real numbers $x$ is the value of the expression

$$
\frac{\left(x+\frac{1}{x}\right)^{3}-\left(x^{3}+\frac{1}{x^{3}}\right)-3}{\left(x+\frac{1}{x}\right)^{2}-\left(x^{2}+\frac{1}{x^{2}}\right)+1}
$$

greater than 1?
## Second Solution.

Let's introduce the substitution $x+\frac{1}{x}=t$. Then we have

$$
\begin{gathered}
x^{3}+\frac{1}{x^{3}}=\left(x+\frac{1}{x}\right)^{3}-3\left(x+\frac{1}{x}\right)=t^{3}-3 t \\
x^{2}+\frac{1}{x^{2}}=\left(x+\frac{1}{x}\right)^{2}-2=t^{2}-2
\end{gathered}
$$

This will transform the given expression into

$$
\frac{\left(x+\frac{1}{x}\right)^{3}-\left(x^{3}+\frac{1}{x^{3}}\right)-3}{\left(x+\frac{1}{x}\right)^{2}-\left(x^{2}+\frac{1}{x^{2}}\right)+1}=\frac{t^{3}-\left(t^{3}-3 t\right)-3}{t^{2}-\left(t^{2}-2\right)+1}=\frac{3 t-3}{3}=t-1=x+\frac{1}{x}-1
$$

The subsequent steps and scoring follow the first solution.