(x^2 - 9)^2 + (x^2 - 2x - 15)^2= 0

To solve the equation \((x^2 - 9)^2 + (x^2 - 2x - 15)^2 = 0\), we need to analyze each term.

Since both terms are squares, for their sum to equal zero, each individual term must also equal zero. Therefore, we can set up the following equations:

1. \( (x^2 - 9)^2 = 0 \)
2. \( (x^2 - 2x - 15)^2 = 0 \)

Now, let’s solve each equation separately.

For the first equation:

\[ (x^2 - 9)^2 = 0 \] Taking the square root of both sides: \[ x^2 - 9 = 0 \] Solving for \(x\): \[ x^2 = 9 \] \[ x = 3 \quad \text{or} \quad x = -3 \]

For the second equation:

\[ (x^2 - 2x - 15)^2 = 0 \] Taking the square root of both sides: \[ x^2 - 2x - 15 = 0 \] This is a quadratic equation. We can solve it by factoring: \[ (x - 5)(x + 3) = 0 \] Setting each factor to zero gives: \[ x - 5 = 0 \Rightarrow x = 5 \] \[ x + 3 = 0 \Rightarrow x = -3 \]

Combining solutions:

Now we combine the solutions from both equations:

• From \((x^2 - 9)^2 = 0\), we have \(x = 3\) and \(x = -3\).
• From \((x^2 - 2x - 15)^2 = 0\), we have \(x = 5\) and \(x = -3\).

The unique solutions for the original equation are: \[ x = 3, \quad x = -3, \quad x = 5 \]

So, the final solution set is: \[ \boxed{3, -3, 5} \]