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Construct a polynomial whose roots are the squares of the roots of the polynomial $x^{3}+x^{2}-2 x-1$.
Let $f(x)=x^{3}+x^{2}-2 x-1$. Since $f(-1)=1$ and $f(0)=-1$, $f(x)$ has three real roots $u, v, w$.

First method. By Vieta's formulas, $\sigma_{1}=u+v+w=-1, \sigma_{2}=u v+u w+v w=-2, \sigma_{3}=u v w=1$. According to the solution of problem $\underline{61030}$

$$
u^{2}+v^{2}+w^{2}=\sigma_{1}^{2}-2 \sigma_{2}=5, u^{2} v^{2}+u^{2} w^{2}+v^{2} w^{2}=\sigma_{2}^{2}-2 \sigma_{1} \sigma_{3}=6, u^{2} v^{2} w^{2}=\sigma_{3}^{2}=1
$$

Therefore, the desired polynomial is $x^{3}-5 x^{2}+6 x-1$.

Second method. Let $g(x)=\left(x-u^{2}\right)\left(x-v^{2}\right)\left(x-w^{2}\right)$. Then

$$
g\left(t^{2}\right)=\left(t^{2}-u^{2}\right)\left(t^{2}-v^{2}\right)\left(t^{2}-w^{2}\right)=(t+u)(t+v)(t+w)(t-u)(t-v)(t-w)=-f(-t) f(t)=
$$

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

## Therefore, $g(x)=x(x-2)^{2}-(x-1)^{2}=x^{3}-4 x^{2}+4 x-\left(x^{2}-2 x+1\right)=x^{3}-5 x^{2}+6 x-1$.

## Answer

$x^{3}-5 x^{2}+6 x-1$.