Resolve the following into partial fractions.
`(x^(2)+5x+7)/((x-3)^(3))`

Answer

To resolve the expression \(\frac{x^2 + 5x + 7}{(x - 3)^3}\) into partial fractions, we will follow these steps:

### Step 1: Set up the partial fraction decomposition
We can express the given fraction as:
\[ \frac{x^2 + 5x + 7}{(x - 3)^3} = \frac{A}{x - 3} + \frac{B}{(x - 3)^2} + \frac{C}{(x - 3)^3} \]
where \(A\), \(B\), and \(C\) are constants that we need to determine.

### Step 2: Clear the denominators
Multiply both sides by \((x - 3)^3\) to eliminate the denominator:
\[ x^2 + 5x + 7 = A(x - 3)^2 + B(x - 3) + C \]

### Step 3: Expand the right-hand side
Now, we expand the right-hand side:
\[ A(x - 3)^2 = A(x^2 - 6x + 9) = Ax^2 - 6Ax + 9A \]
\[ B(x - 3) = Bx - 3B \]
\[ C = C \]
Combining these, we have:
\[ x^2 + 5x + 7 = Ax^2 + (-6A + B)x + (9A - 3B + C) \]

### Step 4: Equate coefficients
Now we equate the coefficients of the corresponding powers of \(x\) from both sides:
1. For \(x^2\): \(A = 1\)
2. For \(x\): \(-6A + B = 5\)
3. For the constant term: \(9A - 3B + C = 7\)

### Step 5: Solve for \(A\), \(B\), and \(C\)
From the first equation, we have:
\[ A = 1 \]
Substituting \(A = 1\) into the second equation:
\[ -6(1) + B = 5 \implies B = 11 \]
Now substituting \(A = 1\) and \(B = 11\) into the third equation:
\[ 9(1) - 3(11) + C = 7 \implies 9 - 33 + C = 7 \implies C = 31 \]

### Step 6: Write the final partial fraction decomposition
Now that we have \(A\), \(B\), and \(C\), we can write the partial fraction decomposition:
\[ \frac{x^2 + 5x + 7}{(x - 3)^3} = \frac{1}{x - 3} + \frac{11}{(x - 3)^2} + \frac{31}{(x - 3)^3} \]

### Final Answer:
\[ \frac{x^2 + 5x + 7}{(x - 3)^3} = \frac{1}{x - 3} + \frac{11}{(x - 3)^2} + \frac{31}{(x - 3)^3} \]