\( \left. \begin{array} { l } { ( x + 2 ) ^ { 2 } + 3 y = x ^ { 3 } - 4 } \\ { 3 x + ( y - \alpha ) ^ { 2 } = y ^ { 2 } } \end{array} \right\} \)

To solve the system of equations given by \[ \begin{cases} (x + 2)^2 + 3y = x^3 - 4 \\ 3x + (y - \alpha)^2 = y^2 \end{cases} \] we will analyze each equation step by step. ### Step 1: Rearranging the first equation The first equation is \[ (x + 2)^2 + 3y = x^3 - 4. \] We can rearrange it to isolate \(y\): \[ 3y = x^3 - 4 - (x + 2)^2. \] Expanding \((x + 2)^2\): \[ (x + 2)^2 = x^2 + 4x + 4. \] Substituting this back into the equation gives: \[ 3y = x^3 - 4 - (x^2 + 4x + 4) = x^3 - x^2 - 4x - 8. \] Now, dividing by 3: \[ y = \frac{x^3 - x^2 - 4x - 8}{3}. \] ### Step 2: Substituting \(y\) into the second equation Now we substitute \(y\) into the second equation: \[ 3x + (y - \alpha)^2 = y^2. \] Substituting for \(y\): \[ 3x + \left(\frac{x^3 - x^2 - 4x - 8}{3} - \alpha\right)^2 = \left(\frac{x^3 - x^2 - 4x - 8}{3}\right)^2. \] This equation is quite complex, so we will simplify it step by step. ### Step 3: Simplifying the second equation Let \(y = \frac{x^3 - x^2 - 4x - 8}{3}\). We need to compute \(y - \alpha\): \[ y - \alpha = \frac{x^3 - x^2 - 4x - 8}{3} - \alpha. \] Now, squaring this term: \[ (y - \alpha)^2 = \left(\frac{x^3 - x^2 - 4x - 8 - 3\alpha}{3}\right)^2 = \frac{(x^3 - x^2 - 4x - 8 - 3\alpha)^2}{9}. \] Now, we also need to compute \(y^2\): \[ y^2 = \left(\frac{x^3 - x^2 - 4x - 8}{3}\right)^2 = \frac{(x^3 - x^2 - 4x - 8)^2}{9}. \] ### Step 4: Setting up the equation Now we can set up the equation: \[ 3x + \frac{(x^3 - x^2 - 4x - 8 - 3\alpha)^2}{9} = \frac{(x^3 - x^2 - 4x - 8)^2}{9}. \] Multiplying through by 9 to eliminate the denominators: \[ 27x + (x^3 - x^2 - 4x - 8 - 3\alpha)^2 = (x^3 - x^2 - 4x - 8)^2. \] ### Step 5: Solving for \(x\) and \(\alpha\) This equation is quite complex and may require numerical methods or specific values for \(\alpha\) to solve. To find specific solutions, we can try substituting values for \(x\) and \(\alpha\) or use numerical methods to find intersections of the two equations. ### Conclusion The system of equations is complex and may not yield simple analytical solutions without specific values for \(\alpha\). If you have a specific value for \(\alpha\) or further constraints, please provide that information for a more targeted solution.