The partial fractions of `((x-1)(x-2)(x-3))/((x-4)(x-5)(x-6))` are …….

Answer

To find the partial fractions of the expression \(\frac{(x-1)(x-2)(x-3)}{(x-4)(x-5)(x-6)}\), we will follow these steps:

### Step 1: Set Up the Partial Fraction Decomposition
We can express the given fraction as a sum of simpler fractions:
\[ \frac{(x-1)(x-2)(x-3)}{(x-4)(x-5)(x-6)} = \frac{A}{x-4} + \frac{B}{x-5} + \frac{C}{x-6} \]
where \(A\), \(B\), and \(C\) are constants that we need to determine.

### Step 2: Clear the Denominator
Multiply both sides of the equation by the denominator \((x-4)(x-5)(x-6)\) to eliminate the fractions:
\[ (x-1)(x-2)(x-3) = A(x-5)(x-6) + B(x-4)(x-6) + C(x-4)(x-5) \]

### Step 3: Expand the Right Side
Now, we will expand the right side:
1. Expand \(A(x-5)(x-6)\):
\[ A(x^2 - 11x + 30) \]
2. Expand \(B(x-4)(x-6)\):
\[ B(x^2 - 10x + 24) \]
3. Expand \(C(x-4)(x-5)\):
\[ C(x^2 - 9x + 20) \]

Combining these, we have:
\[ A(x^2 - 11x + 30) + B(x^2 - 10x + 24) + C(x^2 - 9x + 20) \]

### Step 4: Combine Like Terms
Combine all the terms on the right-hand side:
\[ (A + B + C)x^2 + (-11A - 10B - 9C)x + (30A + 24B + 20C) \]

### Step 5: Set Up the System of Equations
Now, we can equate the coefficients from both sides of the equation:
1. Coefficient of \(x^2\): \(A + B + C = 0\)
2. Coefficient of \(x\): \(-11A - 10B - 9C = 0\)
3. Constant term: \(30A + 24B + 20C = 0\)

### Step 6: Solve the System of Equations
We can solve this system of equations step by step:
1. From \(A + B + C = 0\), we can express \(C\) as \(C = -A - B\).
2. Substitute \(C\) into the second equation:
\[ -11A - 10B - 9(-A - B) = 0 \implies -11A - 10B + 9A + 9B = 0 \implies -2A - B = 0 \implies B = -2A \]
3. Substitute \(B\) and \(C\) into the third equation:
\[ 30A + 24(-2A) + 20(-A - (-2A)) = 0 \implies 30A - 48A - 20A = 0 \implies -38A = 0 \implies A = 0 \]
4. Since \(A = 0\), we find \(B = 0\) and \(C = 0\).

### Step 7: Write the Final Result
Thus, the partial fraction decomposition of \(\frac{(x-1)(x-2)(x-3)}{(x-4)(x-5)(x-6)}\) is:
\[ \frac{0}{x-4} + \frac{0}{x-5} + \frac{0}{x-6} = 0 \]