If `1/(x^(4)+x^(2)+1)=(Ax+B)/(x^(2)+x+1)+(Cx+D)/(x^(2)-x+1) " then "C+D=`

Options

A. -1
B. 1
C. 2
D. 0

Answer

To solve the equation

\[ \frac{1}{x^4 + x^2 + 1} = \frac{Ax + B}{x^2 + x + 1} + \frac{Cx + D}{x^2 - x + 1} \]

we will follow these steps:

### Step 1: Combine the Right-Hand Side
First, we will combine the fractions on the right-hand side. The common denominator will be \((x^2 + x + 1)(x^2 - x + 1)\).

\[ \frac{1}{x^4 + x^2 + 1} = \frac{(Ax + B)(x^2 - x + 1) + (Cx + D)(x^2 + x + 1)}{(x^2 + x + 1)(x^2 - x + 1)} \]

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

1. For \((Ax + B)(x^2 - x + 1)\):
\[ Ax^3 - Ax^2 + Ax + Bx^2 - Bx + B = Ax^3 + (-A + B)x^2 + (A - B)x + B \]

2. For \((Cx + D)(x^2 + x + 1)\):
\[ Cx^3 + Cx^2 + Cx + Dx^2 + Dx + D = Cx^3 + (C + D)x^2 + (C + D)x + D \]

Combining both expansions:
\[ (A + C)x^3 + (-A + B + C + D)x^2 + (A - B + C + D)x + (B + D) \]

### Step 3: Set the Numerator Equal to 1
Since the left-hand side is 1, we can set the numerator equal to 1:
\[ (A + C)x^3 + (-A + B + C + D)x^2 + (A - B + C + D)x + (B + D) = 1 \]

### Step 4: Compare Coefficients
Now we will compare coefficients on both sides:

1. Coefficient of \(x^3\):
\[ A + C = 0 \quad \text{(Equation 1)} \]

2. Coefficient of \(x^2\):
\[ -A + B + C + D = 0 \quad \text{(Equation 2)} \]

3. Coefficient of \(x\):
\[ A - B + C + D = 0 \quad \text{(Equation 3)} \]

4. Constant term:
\[ B + D = 1 \quad \text{(Equation 4)} \]

### Step 5: Solve the Equations
From Equation 1, we have:
\[ C = -A \]

Substituting \(C = -A\) into Equations 2 and 3:

**Equation 2:**
\[ -A + B - A + D = 0 \implies B + D - 2A = 0 \implies B + D = 2A \quad \text{(Equation 5)} \]

**Equation 3:**
\[ A - B - A + D = 0 \implies -B + D = 0 \implies D = B \quad \text{(Equation 6)} \]

### Step 6: Substitute into Equation 4
Now substitute \(D = B\) into Equation 4:
\[ B + B = 1 \implies 2B = 1 \implies B = \frac{1}{2} \]

Then, from Equation 6:
\[ D = \frac{1}{2} \]

### Step 7: Find \(C + D\)
Now we can find \(C\):
From Equation 1:
\[ C = -A \]

Using Equation 5:
\[ B + D = 2A \implies \frac{1}{2} + \frac{1}{2} = 2A \implies 1 = 2A \implies A = \frac{1}{2} \]

Thus,
\[ C = -\frac{1}{2} \]

Finally, we have:
\[ C + D = -\frac{1}{2} + \frac{1}{2} = 0 \]

### Final Answer:
\[ C + D = 0 \]