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

Answer

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

### Step 1: Check the degree of the numerator and denominator
The degree of the numerator \(x^3\) is 3, and the degree of the denominator \((x-1)(x+2)\) is 2. Since the degree of the numerator is greater than the degree of the denominator, we first perform polynomial long division.

### Step 2: Perform polynomial long division
We divide \(x^3\) by \((x-1)(x+2)\):
1. The leading term of the numerator is \(x^3\) and the leading term of the denominator is \(x^2\).
2. Divide \(x^3\) by \(x^2\) to get \(x\).
3. Multiply \(x\) by \((x-1)(x+2)\) to get \(x^3 + x^2 - 2x\).
4. Subtract this from the original numerator:
\[ x^3 - (x^3 + x^2 - 2x) = -x^2 + 2x \]

Now, we have:
\[ \frac{x^3}{(x-1)(x+2)} = x + \frac{-x^2 + 2x}{(x-1)(x+2)} \]

### Step 3: Set up the partial fraction decomposition
Now we need to resolve \(\frac{-x^2 + 2x}{(x-1)(x+2)}\) into partial fractions. We can express it as:
\[ \frac{-x^2 + 2x}{(x-1)(x+2)} = \frac{A}{x-1} + \frac{B}{x+2} \]

### Step 4: Clear the denominators
Multiply both sides by \((x-1)(x+2)\):
\[ -x^2 + 2x = A(x+2) + B(x-1) \]

### Step 5: Expand the right-hand side
Expanding gives:
\[ -x^2 + 2x = Ax + 2A + Bx - B \]
Combining like terms:
\[ -x^2 + 2x = (A + B)x + (2A - B) \]

### Step 6: Set up equations by comparing coefficients
Now, we can set up the following equations by comparing coefficients:
1. For \(x^2\): \(0 = -1\) (no \(x^2\) term on the right)
2. For \(x\): \(2 = A + B\)
3. For the constant term: \(0 = 2A - B\)

### Step 7: Solve the system of equations
From \(2 = A + B\) (1) and \(0 = 2A - B\) (2):
- From (2), we can express \(B\) in terms of \(A\): \(B = 2A\).
- Substitute \(B\) in (1):
\[ 2 = A + 2A \implies 2 = 3A \implies A = \frac{2}{3} \]
- Now substitute \(A\) back to find \(B\):
\[ B = 2A = 2 \cdot \frac{2}{3} = \frac{4}{3} \]

### Step 8: Write the partial fraction decomposition
Now we can write the partial fraction decomposition:
\[ \frac{-x^2 + 2x}{(x-1)(x+2)} = \frac{\frac{2}{3}}{x-1} + \frac{\frac{4}{3}}{x+2} \]

### Final Result
Thus, the complete expression in partial fractions is:
\[ \frac{x^3}{(x-1)(x+2)} = x + \frac{\frac{2}{3}}{x-1} + \frac{\frac{4}{3}}{x+2} \]