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The image presents a problem involving permutations. The permutations are given as follows:

- Permutation \( \alpha \) is defined as:
  \[
  \alpha = 
  \begin{pmatrix}
  1 & 2 & 3 & 4 \\
  3 & 4 & 1 & 2
  \end{pmatrix}
  \]
  which means \( \alpha(1) = 3 \), \( \alpha(2) = 4 \), \( \alpha(3) = 1 \), and \( \alpha(4) = 2 \).

- Permutation \( \beta \) is defined as:
  \[
  \beta = 
  \begin{pmatrix}
  1 & 2 & 3 & 4 \\
  4 & 2 & 1 & 3
  \end{pmatrix}
  \]
  which means \( \beta(1) = 4 \), \( \beta(2) = 2 \), \( \beta(3) = 1 \), and \( \beta(4) = 3 \).

The problem requires solving the equations:

1. \( x \alpha = \beta \)
2. \( \beta y = \alpha^3 \)

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### Step-by-step approach:

1. **Solve the equation \( x \alpha = \beta \):**
   - We need to find the permutation \( x \) such that when it is composed with \( \alpha \), it gives \( \beta \).
   - This requires finding \( x = \beta \alpha^{-1} \), where \( \alpha^{-1} \) is the inverse of \( \alpha \).

2. **Solve the equation \( \beta y = \alpha^3 \):**
   - Here, we are required to find the permutation \( y \) such that when composed with \( \beta \), it gives \( \alpha^3 \).
   - This requires \( y = \beta^{-1} \alpha^3 \), where \( \beta^{-1} \) is the inverse of \( \beta \), and \( \alpha^3 \) is the cube of the permutation \( \alpha \) (computed by applying \( \alpha \) three times).

Let me compute these solutions for you.

Fie permutările α și β date. Rezolvați ecuațiile xα = β și βy = α^3.

Explore how to solve equations involving permutations α and β, such as xα = β and βy = α^3. This problem involves key concepts from group theory and algebra, including permutation inversion and composition.

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