If `(1)/((x-a)(x^(2)+b))=(A)/(x-a)+(Bx+C)/(x^(2)+b)` then `(1)/((x-a)(x^(2)+b)^(2))=`

Options

A. `(A^(2))/(x-a)+(A(Bx+C))/(x^(2)+b)+(Bx+C)/(x^(2)+b)^(2)`
B. `(A^(2))/(x+a)+(A(Bx-C))/(x^(2)+b)+(Bx+C)/((x^(2)-b)^(2))`
C. `(A^(2))/(x+a)+(A(Bx+C))/(x^(2)+b)+(Bx-C)/((x^(2)+b)^(2))`
D. `(A^(2))/(x-a)+(A(Bx-C))/(x^(2)-b)+(Bx+C)/((x^(2)-b)^(2))`

Answer

To solve the equation

\[ \frac{1}{(x-a)(x^2+b)} = \frac{A}{x-a} + \frac{Bx+C}{x^2+b} \]

we will follow these steps:

### Step 1: Combine the Right Side
First, we will combine the right-hand side into a single fraction:

\[ \frac{A}{x-a} + \frac{Bx+C}{x^2+b} = \frac{A(x^2+b) + (Bx+C)(x-a)}{(x-a)(x^2+b)} \]

### Step 2: Expand the Numerator
Next, we will expand the numerator:

\[ A(x^2+b) + (Bx+C)(x-a) = Ax^2 + Ab + Bx^2 - Bax + Cx - Ca \]

Combining like terms:

\[ (A + B)x^2 + (C - Ba)x + (Ab - Ca) \]

### Step 3: Set the Numerators Equal
Now, we equate the numerators from both sides of the equation:

\[ 1 = (A + B)x^2 + (C - Ba)x + (Ab - Ca) \]

### Step 4: Coefficient Comparison
Since the left side is just 1, we can compare coefficients:

1. Coefficient of \(x^2\): \(A + B = 0\)
2. Coefficient of \(x\): \(C - Ba = 0\)
3. Constant term: \(Ab - Ca = 1\)

### Step 5: Solve the System of Equations
From the first equation, we have:

\[ B = -A \]

Substituting \(B = -A\) into the second equation:

\[ C - (-A)a = 0 \implies C + Aa = 0 \implies C = -Aa \]

Substituting \(B = -A\) and \(C = -Aa\) into the third equation:

\[ A(b + Aa) = 1 \]

This gives us:

\[ A(b + Aa) = 1 \implies A = \frac{1}{b + Aa} \]

### Step 6: Finding the Value of \(A\)
To find \(A\), we can rearrange:

\[ A(b + Aa) = 1 \implies A = \frac{1}{b + Aa} \implies A^2a + Ab - 1 = 0 \]

This is a quadratic equation in \(A\). Solving for \(A\) gives us:

\[ A = \frac{-b \pm \sqrt{b^2 + 4a}}{2a} \]

### Step 7: Substitute Back to Find \(B\) and \(C\)
Using the value of \(A\) found, we can substitute back to find \(B\) and \(C\):

\[ B = -A, \quad C = -Aa \]

### Step 8: Final Expression
Now we can express:

\[ \frac{1}{(x-a)(x^2+b)^2} = \frac{A}{(x-a)} + \frac{Bx+C}{(x^2+b)^2} \]

This completes the solution.

---