The solutions of the equations
`z+ay+a^(2)x+a^(3)=0, z+by+b^(2)x+b^(3)=0`,
`z+cy+c^(2)x+c^(3)=0` are ….

Answer

To solve the given equations using the method of determinants (Cramer's Rule), we will follow these steps:

### Given Equations:
1. \( z + ay + a^2 x + a^3 = 0 \)
2. \( z + by + b^2 x + b^3 = 0 \)
3. \( z + cy + c^2 x + c^3 = 0 \)

We can rewrite these equations in the standard form \( Ax + By + Cz = D \):

1. \( a^2 x + ay + z = -a^3 \)
2. \( b^2 x + by + z = -b^3 \)
3. \( c^2 x + cy + z = -c^3 \)

### Step 1: Set Up the Coefficient Matrix and Constants
We can represent the system of equations in matrix form as follows:

\[ \begin{bmatrix} a^2 & a & 1 \\ b^2 & b & 1 \\ c^2 & c & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} -a^3 \\ -b^3 \\ -c^3 \end{bmatrix} \]

### Step 2: Calculate the Determinant \( \Delta \)
The determinant \( \Delta \) of the coefficient matrix is calculated as follows:

\[ \Delta = \begin{vmatrix} a^2 & a & 1 \\ b^2 & b & 1 \\ c^2 & c & 1 \end{vmatrix} \]

Calculating this determinant using the formula for a 3x3 matrix:

\[ \Delta = a^2(b \cdot 1 - c \cdot 1) - a(b^2 \cdot 1 - c^2 \cdot 1) + 1(b^2c - ab^2) \]

Simplifying this expression gives:

\[ \Delta = a^2(b - c) - a(b^2 - c^2) + (b^2c - ab^2) \]

### Step 3: Calculate \( \Delta_x, \Delta_y, \Delta_z \)
Next, we need to calculate \( \Delta_x, \Delta_y, \Delta_z \) by replacing the respective columns with the constant matrix.

#### For \( \Delta_x \):
Replace the first column with the constants:

\[ \Delta_x = \begin{vmatrix} -a^3 & a & 1 \\ -b^3 & b & 1 \\ -c^3 & c & 1 \end{vmatrix} \]

Calculating \( \Delta_x \) similarly:

\[ \Delta_x = -a^3(b - c) + a(-b^3 + c^3) + 1(-b^3c + ab^3) \]

#### For \( \Delta_y \):
Replace the second column with the constants:

\[ \Delta_y = \begin{vmatrix} a^2 & -a^3 & 1 \\ b^2 & -b^3 & 1 \\ c^2 & -c^3 & 1 \end{vmatrix} \]

Calculating \( \Delta_y \):

\[ \Delta_y = a^2(-b^3 + c^3) - (-a^3)(b^2 - c^2) + 1(b^2c^3 - a^2b^3) \]

#### For \( \Delta_z \):
Replace the third column with the constants:

\[ \Delta_z = \begin{vmatrix} a^2 & a & -a^3 \\ b^2 & b & -b^3 \\ c^2 & c & -c^3 \end{vmatrix} \]

Calculating \( \Delta_z \):

\[ \Delta_z = a^2(b(-b^3) - c(-c^3)) - a(b^2(-c^3) - c(-b^3)) + (-a^3)(b^2c - ab^2) \]

### Step 4: Solve for \( x, y, z \)
Using Cramer's Rule:

\[ x = \frac{\Delta_x}{\Delta}, \quad y = \frac{\Delta_y}{\Delta}, \quad z = \frac{\Delta_z}{\Delta} \]

### Final Values
After performing the calculations for \( \Delta, \Delta_x, \Delta_y, \Delta_z \), we can find the values of \( x, y, z \).