Solve the equation `(2b^(2)+x^(2))/(b^(3)-x^(3))-(2x)/(bx+b^(2)+x^(2))+(1)/(x-b)=0` For what value of x is the solution of the equation unique ?

Answer

To solve the equation

\[ \frac{2b^2 + x^2}{b^3 - x^3} - \frac{2x}{bx + b^2 + x^2} + \frac{1}{x - b} = 0 \]

and find the value of \( x \) for which the solution is unique, we will follow these steps:

### Step 1: Rewrite the Equation
We start with the given equation:

\[ \frac{2b^2 + x^2}{b^3 - x^3} - \frac{2x}{bx + b^2 + x^2} + \frac{1}{x - b} = 0 \]

### Step 2: Factor the Denominator
Using the identity for the difference of cubes, \( b^3 - x^3 = (b - x)(b^2 + bx + x^2) \), we can rewrite the first term:

\[ b^3 - x^3 = (b - x)(b^2 + bx + x^2) \]

### Step 3: Find a Common Denominator
The common denominator for the fractions will be:

\[ (b - x)(b^2 + bx + x^2)(x - b) \]

### Step 4: Rewrite Each Term
Now, we can rewrite each term with the common denominator:

1. The first term becomes:
\[ (2b^2 + x^2)(x - b) \]

2. The second term becomes:
\[ -2x(b - x)(b^2 + bx + x^2) \]

3. The third term becomes:
\[ (b^3 - x^3)(1) \]

### Step 5: Combine the Terms
Putting everything together, we have:

\[ (2b^2 + x^2)(x - b) - 2x(b - x)(b^2 + bx + x^2) + (b^3 - x^3) = 0 \]

### Step 6: Expand and Simplify
Now we will expand and simplify the equation. This will lead us to a quadratic equation in terms of \( x \).

### Step 7: Form a Quadratic Equation
After simplification, we arrive at a quadratic equation:

\[ 2x^2 - 3bx + b^2 = 0 \]

### Step 8: Determine Unique Solution
For a quadratic equation \( ax^2 + bx + c = 0 \) to have a unique solution, the discriminant must be zero:

\[ D = b^2 - 4ac \]

In our case:

\[ D = (-3b)^2 - 4 \cdot 2 \cdot b^2 = 9b^2 - 8b^2 = b^2 \]

Setting the discriminant to zero for uniqueness:

\[ b^2 = 0 \implies b = 0 \]

However, since we are looking for the value of \( x \) such that the solution is unique, we must ensure \( b \neq 0 \).

### Step 9: Solve for \( x \)
The solutions to the quadratic equation are given by:

\[ x = \frac{3b \pm \sqrt{b^2}}{2 \cdot 2} = \frac{3b \pm b}{4} \]

This gives two solutions:

1. \( x = b \)
2. \( x = \frac{b}{2} \)

Since \( x = b \) is not valid (as it makes the original equation undefined), the unique solution is:

\[ x = \frac{b}{2} \]

### Conclusion
Thus, the value of \( x \) for which the solution of the equation is unique is:

\[ \boxed{\frac{b}{2}} \]