Let $P (x) = x^{3}-3x+1.$ Find the polynomial $Q$ whose roots are the ﬁfth powers of the roots of $P$.
1. Let \( P(x) = x^3 - 3x + 1 \). We need to find the polynomial \( Q \) whose roots are the fifth powers of the roots of \( P \).
2. Let \( \alpha, \beta, \gamma \) be the roots of \( P(x) \). Then, \( \alpha^3 - 3\alpha + 1 = 0 \), \( \beta^3 - 3\beta + 1 = 0 \), and \( \gamma^3 - 3\gamma + 1 = 0 \).
3. We need to find the polynomial whose roots are \( \alpha^5, \beta^5, \gamma^5 \).
4. Let \( y = x^5 \). Then, \( x = y^{1/5} \).
5. Since \( \alpha, \beta, \gamma \) are roots of \( P(x) \), we have:
   \[
   \alpha^3 = 3\alpha - 1, \quad \beta^3 = 3\beta - 1, \quad \gamma^3 = 3\gamma - 1
   \]
6. We need to express \( \alpha^5, \beta^5, \gamma^5 \) in terms of \( \alpha, \beta, \gamma \).
7. Using the identity \( \alpha^5 = \alpha^2 \cdot \alpha^3 \), we get:
   \[
   \alpha^5 = \alpha^2 (3\alpha - 1) = 3\alpha^3 - \alpha^2 = 3(3\alpha - 1) - \alpha^2 = 9\alpha - 3 - \alpha^2
   \]
8. Similarly, for \( \beta \) and \( \gamma \):
   \[
   \beta^5 = 9\beta - 3 - \beta^2, \quad \gamma^5 = 9\gamma - 3 - \gamma^2
   \]
9. Now, we need to find the polynomial \( Q(y) \) whose roots are \( \alpha^5, \beta^5, \gamma^5 \).
10. The polynomial \( Q(y) \) can be written as:
    \[
    Q(y) = (y - \alpha^5)(y - \beta^5)(y - \gamma^5)
    \]
11. Using the relations \( \alpha^5 = 9\alpha - 3 - \alpha^2 \), \( \beta^5 = 9\beta - 3 - \beta^2 \), and \( \gamma^5 = 9\gamma - 3 - \gamma^2 \), we can express \( Q(y) \) in terms of \( \alpha, \beta, \gamma \).
12. Since \( \alpha, \beta, \gamma \) are roots of \( P(x) \), we know that:
    \[
    \alpha + \beta + \gamma = 0, \quad \alpha\beta + \beta\gamma + \gamma\alpha = -3, \quad \alpha\beta\gamma = -1
    \]
13. Using these symmetric sums, we can find the coefficients of \( Q(y) \).
14. The polynomial \( Q(y) \) is given by:
    \[
    Q(y) = y^3 + 15y^2 - 198y + 1
    \]

The final answer is \( \boxed{y^3 + 15y^2 - 198y + 1} \).