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

Answer

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

### Step 1: Factor the Denominator
The denominator can be factored as follows:
\[ x^3 - 1 = (x-1)(x^2 + x + 1) \]
Thus, we can rewrite the denominator:
\[ (x-1)(x^3 - 1) = (x-1)^2(x^2 + x + 1) \]
So, the expression becomes:
\[ \frac{x^3 + x^2 + 1}{(x-1)^2(x^2 + x + 1)} \]

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

### Step 3: Clear the Denominator
Multiply both sides by the denominator \((x-1)^2(x^2 + x + 1)\):
\[ x^3 + x^2 + 1 = A(x-1)(x^2 + x + 1) + B(x^2 + x + 1) + (Cx + D)(x-1)^2 \]

### Step 4: Expand the Right Side
Now we will expand the right-hand side:
1. Expand \(A(x-1)(x^2 + x + 1)\):
\[ A(x^3 + x^2 + x - x^2 - x - 1) = A(x^3 - 1) \]
2. Expand \(B(x^2 + x + 1)\):
\[ Bx^2 + Bx + B \]
3. Expand \((Cx + D)(x-1)^2\):
\[ (Cx + D)(x^2 - 2x + 1) = Cx^3 - 2Cx^2 + Cx + Dx^2 - 2Dx + D \]

Combining all these, we have:
\[ x^3 + x^2 + 1 = (A + C)x^3 + (-2C + B + D)x^2 + (B + C - 2D)x + (A + B + D) \]

### Step 5: Equate Coefficients
Now we equate the coefficients from both sides:
1. Coefficient of \(x^3\): \(A + C = 1\) (Equation 1)
2. Coefficient of \(x^2\): \(-2C + B + D = 1\) (Equation 2)
3. Coefficient of \(x\): \(B + C - 2D = 0\) (Equation 3)
4. Constant term: \(A + B + D = 1\) (Equation 4)

### Step 6: Solve the Equations
From Equation 1, we can express \(C\) in terms of \(A\):
\[ C = 1 - A \]

Substituting \(C\) into Equations 2, 3, and 4:
1. \(-2(1 - A) + B + D = 1\) simplifies to \(B + D = 3 - 2A\) (Equation 5)
2. \(B + (1 - A) - 2D = 0\) simplifies to \(B - A - 2D = -1\) (Equation 6)
3. \(A + B + D = 1\) (Equation 4 remains unchanged)

Now we have three equations (5, 6, and 4) to solve for \(A\), \(B\), and \(D\).

### Step 7: Solve for Constants
1. From Equation 5: \(D = 3 - 2A - B\)
2. Substitute \(D\) into Equation 6:
\[ B - A - 2(3 - 2A - B) = -1 \]
Simplifying gives:
\[ B - A - 6 + 4A + 2B = -1 \implies 3B + 3A = 5 \implies B + A = \frac{5}{3} \quad (Equation 7) \]
3. Substitute \(B\) from Equation 7 into Equation 4:
\[ A + \frac{5}{3} - A + D = 1 \implies D = 1 - \frac{5}{3} = -\frac{2}{3} \]

Now substituting \(D\) back into Equation 5 gives:
\[ B = 3 - 2A + \frac{2}{3} \implies B = \frac{9}{3} - 2A + \frac{2}{3} = \frac{11}{3} - 2A \]

Substituting \(B\) into Equation 7:
\[ \frac{11}{3} - 2A + A = \frac{5}{3} \implies -A = -2 \implies A = 2 \]
Then substituting \(A\) back to find \(B\) and \(C\):
\[ C = 1 - 2 = -1 \]
\[ B = \frac{11}{3} - 4 = \frac{-1}{3} \]

### Final Values
Thus, we have:
- \(A = 2\)
- \(B = -\frac{1}{3}\)
- \(C = -1\)
- \(D = -\frac{2}{3}\)

### Step 8: Write the Partial Fraction Decomposition
Now we can write the partial fraction decomposition:
\[ \frac{2}{x-1} - \frac{1/3}{(x-1)^2} - \frac{x/3 + 2/3}{x^2 + x + 1} \]

### Final Answer
The final answer is:
\[ \frac{2}{x-1} - \frac{1/3}{(x-1)^2} - \frac{x + 2}{3(x^2 + x + 1)} \]