Resolve `(3x^(3)-2x^(2)-1)/(x^(4)+x^(2)+1)` into partial fractions.

Answer

To resolve the expression \(\frac{3x^3 - 2x^2 - 1}{x^4 + x^2 + 1}\) into partial fractions, we will follow these steps:

### Step 1: Factor the Denominator
The denominator is \(x^4 + x^2 + 1\). We can rewrite it as:
\[ x^4 + x^2 + 1 = (x^2 + 1)^2 - x^2 = (x^2 + 1 + x)(x^2 + 1 - x) \]
Thus, we can express the denominator as:
\[ x^4 + x^2 + 1 = (x^2 - x + 1)(x^2 + x + 1) \]

### Step 2: Set Up Partial Fraction Decomposition
We can express the fraction as:
\[ \frac{3x^3 - 2x^2 - 1}{(x^2 - x + 1)(x^2 + x + 1)} = \frac{Ax + B}{x^2 - x + 1} + \frac{Cx + D}{x^2 + x + 1} \]
where \(A\), \(B\), \(C\), and \(D\) are constants that we need to determine.

### Step 3: Combine the Right Side
To combine the right side, we will find a common denominator:
\[ \frac{Ax + B}{x^2 - x + 1} + \frac{Cx + D}{x^2 + x + 1} = \frac{(Ax + B)(x^2 + x + 1) + (Cx + D)(x^2 - x + 1)}{(x^2 - x + 1)(x^2 + x + 1)} \]

### Step 4: Expand the Numerator
Expanding the numerator gives:
\[ (Ax + B)(x^2 + x + 1) = Ax^3 + Ax^2 + Ax + Bx^2 + Bx + B \]
\[ (Cx + D)(x^2 - x + 1) = Cx^3 - Cx^2 + Cx + Dx^2 - Dx + D \]
Combining these, we have:
\[ (A + C)x^3 + (A + B - C + D)x^2 + (A + B + C - D)x + (B + D) \]

### Step 5: Set Up the Equation
Now, we equate the coefficients of the numerator:
\[ 3x^3 - 2x^2 - 1 = (A + C)x^3 + (A + B - C + D)x^2 + (A + B + C - D)x + (B + D) \]
This gives us the following system of equations:
1. \(A + C = 3\) (coefficient of \(x^3\))
2. \(A + B - C + D = -2\) (coefficient of \(x^2\))
3. \(A + B + C - D = 0\) (coefficient of \(x\))
4. \(B + D = -1\) (constant term)

### Step 6: Solve the System of Equations
From equation 1, we can express \(C\) as:
\[ C = 3 - A \]
Substituting \(C\) into the other equations:
- From equation 2:
\[ A + B - (3 - A) + D = -2 \implies 2A + B + D - 3 = -2 \implies 2A + B + D = 1 \quad (5) \]
- From equation 3:
\[ A + B + (3 - A) - D = 0 \implies B + 3 - D = 0 \implies B - D = -3 \quad (6) \]
- From equation 4:
\[ B + D = -1 \quad (4) \]

Now we can solve equations (6) and (4):
From (6):
\[ B = D - 3 \]
Substituting into (4):
\[ (D - 3) + D = -1 \implies 2D - 3 = -1 \implies 2D = 2 \implies D = 1 \]
Substituting \(D = 1\) back into (4):
\[ B + 1 = -1 \implies B = -2 \]
Substituting \(B = -2\) into (6):
\[ -2 - D = -3 \implies D = 1 \]
Now substituting \(B\) and \(D\) into (5):
\[ 2A - 2 + 1 = 1 \implies 2A - 1 = 1 \implies 2A = 2 \implies A = 1 \]
Finally, substituting \(A = 1\) into (1):
\[ 1 + C = 3 \implies C = 2 \]

### Step 7: Write the Final Result
Thus, we have:
\[ A = 1, B = -2, C = 2, D = 1 \]
The partial fraction decomposition is:
\[ \frac{3x^3 - 2x^2 - 1}{x^4 + x^2 + 1} = \frac{1x - 2}{x^2 - x + 1} + \frac{2x + 1}{x^2 + x + 1} \]