The partial fractions of `(x^(2)-5x+7)/((x-1)^3)` are ………

Answer

To find the partial fractions of the expression \(\frac{x^2 - 5x + 7}{(x-1)^3}\), we will follow these steps:

### Step 1: Set up the partial fraction decomposition

Since the denominator is \((x-1)^3\), we can express the fraction in terms of its partial fractions as follows:

\[ \frac{x^2 - 5x + 7}{(x-1)^3} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{(x-1)^3} \]

### Step 2: Combine the right-hand side over a common denominator

To combine the right-hand side, we need to express it with a common denominator of \((x-1)^3\):

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

### Step 3: Set the numerators equal

Now, we can set the numerators equal to each other:

\[ x^2 - 5x + 7 = A(x-1)^2 + B(x-1) + C \]

### Step 4: Expand the right-hand side

Expanding \(A(x-1)^2\):

\[ A(x^2 - 2x + 1) = Ax^2 - 2Ax + A \]

Expanding \(B(x-1)\):

\[ B(x - 1) = Bx - B \]

Now, combine these expansions:

\[ Ax^2 - 2Ax + A + Bx - B + C = Ax^2 + (-2A + B)x + (A - B + C) \]

### Step 5: Compare coefficients

Now, we can compare the coefficients from both sides of the equation:

1. Coefficient of \(x^2\): \(A = 1\)
2. Coefficient of \(x\): \(-2A + B = -5\)
3. Constant term: \(A - B + C = 7\)

### Step 6: Solve the system of equations

From \(A = 1\), we substitute into the second equation:

\[ -2(1) + B = -5 \implies B = -5 + 2 = -3 \]

Now substitute \(A\) and \(B\) into the third equation:

\[ 1 - (-3) + C = 7 \implies 1 + 3 + C = 7 \implies C = 7 - 4 = 3 \]

### Step 7: Write the final partial fraction decomposition

Now that we have \(A\), \(B\), and \(C\):

\[ A = 1, \quad B = -3, \quad C = 3 \]

Thus, the partial fraction decomposition is:

\[ \frac{x^2 - 5x + 7}{(x-1)^3} = \frac{1}{x-1} - \frac{3}{(x-1)^2} + \frac{3}{(x-1)^3} \]

### Final Answer

The partial fractions of \(\frac{x^2 - 5x + 7}{(x-1)^3}\) are:

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

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