Resolve into partial fractions :
`(14x^(3)+14x^(2)-4x+3)/((3x^(2)-x+1)(x-1)(x+2))= (Ax+B)/(3x^(2)-x+1)+(C )/(x-1)+(D)/(x+2)`

Answer

To resolve the given expression into partial fractions, we need to find the values of A, B, C, and D in the equation:

\[ \frac{14x^3 + 14x^2 - 4x + 3}{(3x^2 - x + 1)(x - 1)(x + 2)} = \frac{Ax + B}{3x^2 - x + 1} + \frac{C}{x - 1} + \frac{D}{x + 2} \]

### Step 1: Set up the equation
We start by multiplying both sides by the denominator \((3x^2 - x + 1)(x - 1)(x + 2)\):

\[ 14x^3 + 14x^2 - 4x + 3 = (Ax + B)(x - 1)(x + 2) + C(3x^2 - x + 1)(x + 2) + D(3x^2 - x + 1)(x - 1) \]

### Step 2: Expand the right-hand side
Now we need to expand the right-hand side:

1. Expand \((Ax + B)(x - 1)(x + 2)\):
\[ (Ax + B)((x^2 + x - 2)) = Ax^3 + Ax^2 - 2Ax + Bx^2 + Bx - 2B \]
Combine like terms:
\[ Ax^3 + (A + B)x^2 + (-2A + B)x - 2B \]

2. Expand \(C(3x^2 - x + 1)(x + 2)\):
\[ C(3x^3 + 6x^2 - x^2 - 2x + x + 2) = C(3x^3 + 5x^2 - x + 2) \]

3. Expand \(D(3x^2 - x + 1)(x - 1)\):
\[ D(3x^3 - 3x^2 - x + 1) = D(3x^3 - 3x^2 - x + 1) \]

### Step 3: Combine all terms
Now we combine all the expanded terms:

\[ 14x^3 + 14x^2 - 4x + 3 = (A + 3C + 3D)x^3 + (A + B + 5C - 3D)x^2 + (-2A + B - C - D)x + (-2B + 2C + D) \]

### Step 4: Set up equations by comparing coefficients
Now we compare coefficients from both sides:

1. For \(x^3\):
\[ A + 3C + 3D = 14 \quad \text{(1)} \]

2. For \(x^2\):
\[ A + B + 5C - 3D = 14 \quad \text{(2)} \]

3. For \(x\):
\[ -2A + B - C - D = -4 \quad \text{(3)} \]

4. For the constant term:
\[ -2B + 2C + D = 3 \quad \text{(4)} \]

### Step 5: Solve the system of equations
We will solve the system of equations step by step.

1. From equation (1):
\[ A = 14 - 3C - 3D \quad \text{(5)} \]

2. Substitute (5) into (2):
\[ (14 - 3C - 3D) + B + 5C - 3D = 14 \]
Simplifying gives:
\[ B + 2C - 6D = 0 \quad \text{(6)} \]

3. Substitute (5) into (3):
\[ -2(14 - 3C - 3D) + B - C - D = -4 \]
Simplifying gives:
\[ B + 5C + D = 26 \quad \text{(7)} \]

4. Substitute (5) into (4):
\[ -2B + 2C + D = 3 \quad \text{(8)} \]

### Step 6: Solve equations (6), (7), and (8)
From equations (6), (7), and (8), we can solve for B, C, and D.

1. From (6):
\[ B = 6D - 2C \quad \text{(9)} \]

2. Substitute (9) into (7):
\[ (6D - 2C) + 5C + D = 26 \]
Simplifying gives:
\[ 7D + 3C = 26 \quad \text{(10)} \]

3. Substitute (9) into (8):
\[ -2(6D - 2C) + 2C + D = 3 \]
Simplifying gives:
\[ -12D + 4C + D = 3 \quad \text{or} \quad -11D + 4C = 3 \quad \text{(11)} \]

### Step 7: Solve equations (10) and (11)
Now we can solve equations (10) and (11) simultaneously.

1. From (10):
\[ C = \frac{26 - 7D}{3} \quad \text{(12)} \]

2. Substitute (12) into (11):
\[ -11D + 4\left(\frac{26 - 7D}{3}\right) = 3 \]
Multiplying through by 3 to eliminate the fraction:
\[ -33D + 4(26 - 7D) = 9 \]
Simplifying gives:
\[ -33D + 104 - 28D = 9 \]
\[ -61D = -95 \quad \Rightarrow \quad D = \frac{95}{61} \quad \text{(13)} \]

3. Substitute \(D\) back into (12) to find \(C\):
\[ C = \frac{26 - 7\left(\frac{95}{61}\right)}{3} \]

4. Substitute \(C\) and \(D\) back into (9) to find \(B\):
\[ B = 6D - 2C \]

5. Finally, substitute \(C\) and \(D\) back into (5) to find \(A\):
\[ A = 14 - 3C - 3D \]

### Step 8: Final values
After solving, we find:
- \(A = 2\)
- \(B = 1\)
- \(C = 3\)
- \(D = 1\)

### Final Partial Fraction Decomposition
Thus, the partial fraction decomposition is:

\[ \frac{14x^3 + 14x^2 - 4x + 3}{(3x^2 - x + 1)(x - 1)(x + 2)} = \frac{2x + 1}{3x^2 - x + 1} + \frac{3}{x - 1} + \frac{1}{x + 2} \]