11. (SWE 6) Let $P, Q, R$ be polynomials and let $S(x)=P\left(x^{3}\right)+x Q\left(x^{3}\right)+$ $x^{2} R\left(x^{3}\right)$ be a polynomial of degree $n$ whose roots $x_{1}, \ldots, x_{n}$ are distinct. Construct with the aid of the polynomials $P, Q, R$ a polynomial $T$ of degree $n$ that has the roots $x_{1}^{3}, x_{2}^{3}, \ldots, x_{n}^{3}$ -
11. Let $S(x)=\left(x-x_{1}\right)\left(x-x_{2}\right) \cdots\left(x-x_{n}\right)$. We have $x^{3}-x_{i}^{3}=\left(x-x_{i}\right)(\omega x-$ $\left.x_{1}\right)\left(\omega^{2} x-x_{i}\right)$, where $\omega$ is a primitive third root of 1 . Multiplying these equalities for $i=1, \ldots, n$ we obtain
$$
T\left(x^{3}\right)=\left(x^{3}-x_{1}^{3}\right)\left(x^{3}-x_{2}^{3}\right) \cdots\left(x^{3}-x_{n}^{3}\right)=S(x) S(\omega x) S\left(\omega^{2} x\right) .
$$

Since $S(\omega x)=P\left(x^{3}\right)+\omega x Q\left(x^{3}\right)+\omega^{2} x^{2} R\left(x^{3}\right)$ and $S\left(\omega^{2} x\right)=P\left(x^{3}\right)+$ $\omega^{2} x Q\left(x^{3}\right)+\omega x^{2} R\left(x^{3}\right)$, the above expression reduces to
$$
T\left(x^{3}\right)=P^{3}\left(x^{3}\right)+x^{3} Q^{3}\left(x^{3}\right)+x^{6} R^{3}\left(x^{3}\right)-3 P\left(x^{3}\right) Q\left(x^{3}\right) R\left(x^{3}\right) .
$$

Therefore the zeros of the polynomial
$$
T(x)=P^{3}(x)+x Q^{3}(x)+x^{2} R^{3}(x)-3 P(x) Q(x) R(x)
$$
are exactly $x_{1}^{3}, \ldots, x_{n}^{3}$. It is easily verified that $\operatorname{deg} T=\operatorname{deg} S=n$, and hence $T$ is the desired polynomial.