Resolve into Partial Fractions
(ii) `(x^(3))/((2x-1)(x+2)(x-3))`

Answer

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

### Step 1: Set up the partial fraction decomposition
We express the fraction as:
\[ \frac{x^3}{(2x-1)(x+2)(x-3)} = \frac{A}{2x-1} + \frac{B}{x+2} + \frac{C}{x-3} \]
where \(A\), \(B\), and \(C\) are constants that we need to determine.

### Step 2: Clear the denominators
Multiply both sides by the denominator \((2x-1)(x+2)(x-3)\) to eliminate the fractions:
\[ x^3 = A(x+2)(x-3) + B(2x-1)(x-3) + C(2x-1)(x+2) \]

### Step 3: Expand the right-hand side
Now we need to expand the right-hand side:
1. For \(A(x+2)(x-3)\):
\[ A[(x^2 - 3x + 2x - 6)] = A[x^2 - x - 6] \]
2. For \(B(2x-1)(x-3)\):
\[ B[(2x^2 - 6x - x + 3)] = B[2x^2 - 7x + 3] \]
3. For \(C(2x-1)(x+2)\):
\[ C[(2x^2 + 4x - x - 2)] = C[2x^2 + 3x - 2] \]

Combining these, we have:
\[ x^3 = A(x^2 - x - 6) + B(2x^2 - 7x + 3) + C(2x^2 + 3x - 2) \]

### Step 4: Collect like terms
Now, we can collect like terms on the right-hand side:
\[ x^3 = (A + 2B + 2C)x^2 + (-A - 7B + 3C)x + (-6A + 3B - 2C) \]

### Step 5: Set up a system of equations
Now we can equate the coefficients from both sides:
1. Coefficient of \(x^3\): \(0 = 1\) (not applicable)
2. Coefficient of \(x^2\): \(A + 2B + 2C = 0\)
3. Coefficient of \(x\): \(-A - 7B + 3C = 0\)
4. Constant term: \(-6A + 3B - 2C = 0\)

### Step 6: Solve the system of equations
We can solve the equations step by step:

1. From \(A + 2B + 2C = 0\) (1)
2. From \(-A - 7B + 3C = 0\) (2)
3. From \(-6A + 3B - 2C = 0\) (3)

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

Substituting \(A\) in (2):
\[ -(-2B - 2C) - 7B + 3C = 0 \implies 2B + 2C - 7B + 3C = 0 \implies -5B + 5C = 0 \implies B = C \]

Substituting \(B = C\) into (1):
\[ A + 2B + 2B = 0 \implies A + 4B = 0 \implies A = -4B \]

Substituting \(A\) and \(B\) into (3):
\[ -6(-4B) + 3B - 2B = 0 \implies 24B + 3B - 2B = 0 \implies 25B = 0 \implies B = 0 \]

Thus, \(B = 0\) implies \(C = 0\) and \(A = 0\).

### Step 7: Final values
Now substituting back, we find:
- \(A = 0\)
- \(B = 0\)
- \(C = 0\)

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

### Conclusion
The final result is:
\[ \frac{x^3}{(2x-1)(x+2)(x-3)} = 0 \]