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

Answer

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

### Step 1: Set up the Partial Fraction Decomposition
We start by expressing \(\frac{1}{x^3(x+2)}\) in terms of its partial fractions. The form will be:

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

### Step 2: Multiply through by the Denominator
Next, we multiply both sides by the denominator \(x^3(x+2)\) to eliminate the fractions:

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

### Step 3: Expand the Right Side
Now, we will expand the right-hand side:

\[ 1 = A x^3 + 2A x^2 + B x^2 + 2B x + C x + 2C + D x^3 \]

Combining like terms gives:

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

### Step 4: Set Up a System of Equations
Now, we will equate the coefficients of the corresponding powers of \(x\) on both sides:

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

### Step 5: Solve the System of Equations
From the fourth equation, we can solve for \(C\):

\[ 2C = 1 \implies C = \frac{1}{2} \]

Substituting \(C\) into the third equation:

\[ 2B + \frac{1}{2} = 0 \implies 2B = -\frac{1}{2} \implies B = -\frac{1}{4} \]

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

\[ 2A - \frac{1}{4} = 0 \implies 2A = \frac{1}{4} \implies A = \frac{1}{8} \]

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

\[ \frac{1}{8} + D = 0 \implies D = -\frac{1}{8} \]

### Step 6: Write the Partial Fraction Decomposition
Now that we have the values of \(A\), \(B\), \(C\), and \(D\), we can write the partial fraction decomposition:

\[ \frac{1}{x^3(x+2)} = \frac{1/8}{x} - \frac{1/4}{x^2} + \frac{1/2}{x^3} - \frac{1/8}{x+2} \]

### Final Answer
Thus, the resolved form into partial fractions is:

\[ \frac{1}{x^3(x+2)} = \frac{1/8}{x} - \frac{1/4}{x^2} + \frac{1/2}{x^3} - \frac{1/8}{x+2} \]

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