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

• Permutation $\alpha$ is defined as: \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$$.
• Permutation $\beta$ is defined as: \begin{pmatrix} 1 & 2 & 3 & 4 \\ 4 & 2 & 1 & 3 \end{pmatrix}$$ 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. $\beta y = \alpha^3$

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.