Resolve into Partial Fractions
(ix) `(3x-18)/(x^(3)(x+3))`

Answer

To resolve the expression \(\frac{3x - 18}{x^3(x + 3)}\) into 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 its partial fractions. Since the denominator \(x^3(x + 3)\) consists of a cubic term and a linear term, we can express it as:

\[ \frac{3x - 18}{x^3(x + 3)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x^3} + \frac{D}{x + 3} \]

### Step 2: Combine the Right Side
Next, we will combine the right-hand side over a common denominator:

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

### Step 3: Expand the Numerator
Now, we will expand the numerator:

\[ A(x^2)(x + 3) = Ax^3 + 3Ax^2 \]
\[ B(x)(x + 3) = Bx^2 + 3Bx \]
\[ C(x + 3) = Cx + 3C \]
\[ D(x^3) = Dx^3 \]

Combining these gives:

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

### Step 4: Set Up the Equation
Now, we equate the numerators from both sides:

\[ 3x - 18 = (A + D)x^3 + (3A + B)x^2 + (3B + C)x + 3C \]

### Step 5: Compare Coefficients
We can now compare coefficients of the powers of \(x\):

1. Coefficient of \(x^3\): \(A + D = 0\)
2. Coefficient of \(x^2\): \(3A + B = 0\)
3. Coefficient of \(x^1\): \(3B + C = 3\)
4. Constant term: \(3C = -18\)

### Step 6: Solve the System of Equations
From the fourth equation, we can solve for \(C\):

\[ 3C = -18 \implies C = -6 \]

Substituting \(C = -6\) into the third equation:

\[ 3B - 6 = 3 \implies 3B = 9 \implies B = 3 \]

Now substituting \(B = 3\) into the second equation:

\[ 3A + 3 = 0 \implies 3A = -3 \implies A = -1 \]

Finally, substituting \(A = -1\) into the first equation:

\[ -1 + D = 0 \implies D = 1 \]

### Step 7: Write the Partial Fraction Decomposition
We have found:

- \(A = -1\)
- \(B = 3\)
- \(C = -6\)
- \(D = 1\)

Thus, the partial fraction decomposition is:

\[ \frac{3x - 18}{x^3(x + 3)} = \frac{-1}{x} + \frac{3}{x^2} + \frac{-6}{x^3} + \frac{1}{x + 3} \]

### Final Answer
The final result is:

\[ \frac{3x - 18}{x^3(x + 3)} = \frac{-1}{x} + \frac{3}{x^2} - \frac{6}{x^3} + \frac{1}{x + 3} \]

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