If `x^(2) + y^(2) + z^(2) + 12x + 4y + 5=0` find `x^(12) + y+ z^(30)`= ?

Answer

To solve the equation \( x^2 + y^2 + z^2 + 12x + 4y + 5 = 0 \) and find \( x^{12} + y + z^{30} \), we will follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ x^2 + y^2 + z^2 + 12x + 4y + 5 = 0 \]
We can rearrange it to group the terms involving \( x \) and \( y \):
\[ x^2 + 12x + y^2 + 4y + z^2 + 5 = 0 \]

### Step 2: Complete the square for \( x \) and \( y \)
Next, we will complete the square for the \( x \) and \( y \) terms.

For \( x^2 + 12x \):
\[ x^2 + 12x = (x + 6)^2 - 36 \]

For \( y^2 + 4y \):
\[ y^2 + 4y = (y + 2)^2 - 4 \]

Substituting these back into the equation gives:
\[ (x + 6)^2 - 36 + (y + 2)^2 - 4 + z^2 + 5 = 0 \]
Simplifying this, we have:
\[ (x + 6)^2 + (y + 2)^2 + z^2 - 35 = 0 \]
Thus,
\[ (x + 6)^2 + (y + 2)^2 + z^2 = 35 \]

### Step 3: Analyze the equation
The equation \( (x + 6)^2 + (y + 2)^2 + z^2 = 35 \) represents a sphere centered at \( (-6, -2, 0) \) with a radius of \( \sqrt{35} \).

### Step 4: Find integer solutions
To find integer solutions, we can try different values for \( x, y, z \). Let's try \( x = -1 \), \( y = 1 \), and \( z = 1 \):
\[ (-1 + 6)^2 + (1 + 2)^2 + (1)^2 = 5^2 + 3^2 + 1^2 = 25 + 9 + 1 = 35 \]
This satisfies the equation.

### Step 5: Substitute values into \( x^{12} + y + z^{30} \)
Now we substitute \( x = -1 \), \( y = 1 \), and \( z = 1 \) into the expression \( x^{12} + y + z^{30} \):
\[ x^{12} + y + z^{30} = (-1)^{12} + 1 + 1^{30} \]
Calculating each term:
\[ (-1)^{12} = 1, \quad 1 = 1, \quad 1^{30} = 1 \]
Thus,
\[ x^{12} + y + z^{30} = 1 + 1 + 1 = 3 \]

### Final Answer
The final answer is:
\[ \boxed{3} \]