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 \( S(x) = (x - x_1)(x - x_2) \cdots (x - x_n) \). We have \( x^3 - x_i^3 = (x - x_i)(\omega x - x_i)(\omega^2 x - x_i) \), where \( \omega \) is a primitive third root of 1. Multiplying these equalities for \( i = 1, \ldots, n \) we obtain 
\[ T(x^3) = (x^3 - x_1^3)(x^3 - x_2^3) \cdots (x^3 - x_n^3) = S(x) S(\omega x) S(\omega^2 x) \]
Since \( S(\omega x) = P(x^3) + \omega x Q(x^3) + \omega^2 x^2 R(x^3) \) and \( S(\omega^2 x) = P(x^3) + \omega^2 x Q(x^3) + \omega x^2 R(x^3) \), the above expression reduces to 
\[ T(x^3) = P^3(x^3) + x^3 Q^3(x^3) + x^6 R^3(x^3) - 3 P(x^3) Q(x^3) R(x^3) \]
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.