Solve `(x^(3))/((2x-1)(x-1)^(2))`

Answer

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

### Step 1: Set Up the Partial Fraction Decomposition
We start by expressing the given fraction in terms of partial fractions. The denominators are \(2x - 1\) and \((x - 1)^2\). Thus, we can write:

\[ \frac{x^3}{(2x-1)(x-1)^2} = \frac{A}{2x-1} + \frac{B}{x-1} + \frac{C}{(x-1)^2} \]

where \(A\), \(B\), and \(C\) are constants that we need to determine.

### Step 2: Multiply Through by the Denominator
To eliminate the denominators, we multiply both sides of the equation by \((2x-1)(x-1)^2\):

\[ x^3 = A(x-1)^2 + B(2x-1)(x-1) + C(2x-1) \]

### Step 3: Expand the Right Side
Now we expand the right side:

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

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

3. For \(C(2x-1)\):
\[ C(2x - 1) \]

Combining these, we have:

\[ x^3 = (A + 2B)x^2 + (-2A - 3B + 2C)x + (A + B - C) \]

### Step 4: Equate Coefficients
Now we equate the coefficients from both sides of the equation:

1. Coefficient of \(x^3\): \(1 = 0\) (no \(x^3\) term on the right side)
2. Coefficient of \(x^2\): \(0 = A + 2B\)
3. Coefficient of \(x\): \(0 = -2A - 3B + 2C\)
4. Constant term: \(0 = A + B - C\)

### Step 5: Solve the System of Equations
From the equations we derived:

1. \(A + 2B = 0\) (Equation 1)
2. \(-2A - 3B + 2C = 0\) (Equation 2)
3. \(A + B - C = 0\) (Equation 3)

From Equation 1, we can express \(A\) in terms of \(B\):
\[ A = -2B \]

Substituting \(A = -2B\) into Equation 2:
\[ -2(-2B) - 3B + 2C = 0 \implies 4B - 3B + 2C = 0 \implies B + 2C = 0 \implies C = -\frac{B}{2} \]

Now substitute \(A = -2B\) and \(C = -\frac{B}{2}\) into Equation 3:
\[ -2B + B + \frac{B}{2} = 0 \]
Multiplying through by 2 to eliminate the fraction:
\[ -4B + 2B + B = 0 \implies -B = 0 \implies B = 0 \]

Thus, substituting \(B = 0\) back:
\[ A = -2(0) = 0, \quad C = -\frac{0}{2} = 0 \]

### Step 6: Conclusion
The values of \(A\), \(B\), and \(C\) are:
\[ A = 0, \quad B = 0, \quad C = 0 \]

Thus, the partial fraction decomposition is:
\[ \frac{x^3}{(2x-1)(x-1)^2} = 0 \]