Let $\alpha,\beta,\gamma$ be the three real roots of the polynomial $x^3-x^2-2x+1=0$. Find all possible values of $\tfrac{\alpha}{\beta}+\tfrac{\beta}{\gamma}+\tfrac{\gamma}{\alpha}$.
1. **Using Vieta's Theorem:**
   Given the polynomial \(x^3 - x^2 - 2x + 1 = 0\), let \(\alpha, \beta, \gamma\) be its roots. By Vieta's formulas, we have:
   \[
   \alpha + \beta + \gamma = 1,
   \]
   \[
   \alpha\beta + \beta\gamma + \gamma\alpha = -2,
   \]
   \[
   \alpha\beta\gamma = -1.
   \]

2. **Using the identity for the sum of cubes:**
   The identity for the sum of cubes of the roots of a polynomial is:
   \[
   \alpha^3 + \beta^3 + \gamma^3 - 3\alpha\beta\gamma = (\alpha + \beta + \gamma)((\alpha + \beta + \gamma)^2 - 3(\alpha\beta + \beta\gamma + \gamma\alpha)).
   \]
   Substituting the values from Vieta's formulas:
   \[
   \alpha^3 + \beta^3 + \gamma^3 - 3(-1) = 1(1^2 - 3(-2)),
   \]
   \[
   \alpha^3 + \beta^3 + \gamma^3 + 3 = 1(1 + 6),
   \]
   \[
   \alpha^3 + \beta^3 + \gamma^3 + 3 = 7,
   \]
   \[
   \alpha^3 + \beta^3 + \gamma^3 = 4.
   \]

3. **Defining \(\sigma_1\) and \(\sigma_2\):**
   Let:
   \[
   \sigma_1 = \frac{\alpha}{\beta} + \frac{\beta}{\gamma} + \frac{\gamma}{\alpha},
   \]
   \[
   \sigma_2 = \frac{\beta}{\alpha} + \frac{\gamma}{\beta} + \frac{\alpha}{\gamma}.
   \]

4. **Finding \(\sigma_1 + \sigma_2\):**
   Using the identity:
   \[
   \sigma_1 + \sigma_2 = (\alpha + \beta + \gamma)\left(\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma}\right) - 3.
   \]
   Since:
   \[
   \frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} = \frac{\beta\gamma + \gamma\alpha + \alpha\beta}{\alpha\beta\gamma} = \frac{-2}{-1} = 2,
   \]
   we get:
   \[
   \sigma_1 + \sigma_2 = 1 \cdot 2 - 3 = -1.
   \]

5. **Finding \(\sigma_1 \sigma_2\):**
   Using the identity:
   \[
   \sigma_1 \sigma_2 = 3 + 3\left(\frac{\alpha^3 + \beta^3 + \gamma^3}{\alpha\beta\gamma}\right) + \alpha\beta\gamma\left(\frac{1}{\alpha^3} + \frac{1}{\beta^3} + \frac{1}{\gamma^3}\right).
   \]
   We already know:
   \[
   \frac{\alpha^3 + \beta^3 + \gamma^3}{\alpha\beta\gamma} = \frac{4}{-1} = -4.
   \]
   To find \(\frac{1}{\alpha^3} + \frac{1}{\beta^3} + \frac{1}{\gamma^3}\), consider the polynomial with roots \(\frac{1}{\alpha}, \frac{1}{\beta}, \frac{1}{\gamma}\):
   \[
   x^3 - 2x^2 - x + 1 = 0.
   \]
   Using the identity:
   \[
   \frac{1}{\alpha^3} + \frac{1}{\beta^3} + \frac{1}{\gamma^3} - \frac{3}{\alpha\beta\gamma} = \left(\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma}\right)\left(\left(\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma}\right)^2 - 3\left(\frac{1}{\alpha\beta} + \frac{1}{\beta\gamma} + \frac{1}{\gamma\alpha}\right)\right).
   \]
   Substituting the values:
   \[
   \frac{1}{\alpha^3} + \frac{1}{\beta^3} + \frac{1}{\gamma^3} - 3(-1) = 2(2^2 - 3(-2)),
   \]
   \[
   \frac{1}{\alpha^3} + \frac{1}{\beta^3} + \frac{1}{\gamma^3} + 3 = 2(4 + 6),
   \]
   \[
   \frac{1}{\alpha^3} + \frac{1}{\beta^3} + \frac{1}{\gamma^3} + 3 = 20,
   \]
   \[
   \frac{1}{\alpha^3} + \frac{1}{\beta^3} + \frac{1}{\gamma^3} = 17.
   \]
   Therefore:
   \[
   \sigma_1 \sigma_2 = 3 + 3(-4) + (-1)(17) = 3 - 12 - 17 = -26.
   \]

6. **Solving the quadratic equation:**
   \(\sigma_1\) and \(\sigma_2\) are the roots of the quadratic equation:
   \[
   y^2 + y - 12 = 0.
   \]
   Solving this quadratic equation:
   \[
   y = \frac{-1 \pm \sqrt{1 + 48}}{2} = \frac{-1 \pm 7}{2}.
   \]
   Thus, the roots are:
   \[
   y = 3 \quad \text{and} \quad y = -4.
   \]

The final answer is \(\boxed{3, -4}\).