Solve for x : `(1)/((x-1)(x-2))+(1)/((x-2)(x-3))+(1)/((x-3)(x-4))=(1)/(6)`

Answer

To solve the equation

\[ \frac{1}{(x-1)(x-2)} + \frac{1}{(x-2)(x-3)} + \frac{1}{(x-3)(x-4)} = \frac{1}{6}, \]

we will follow these steps:

### Step 1: Combine the fractions on the left-hand side
We can rewrite the left-hand side by finding a common denominator, which is \((x-1)(x-2)(x-3)(x-4)\).

The first term becomes:
\[ \frac{(x-3)(x-4)}{(x-1)(x-2)(x-3)(x-4)}. \]

The second term becomes:
\[ \frac{(x-1)(x-4)}{(x-2)(x-3)(x-1)(x-4)}. \]

The third term becomes:
\[ \frac{(x-1)(x-2)}{(x-3)(x-4)(x-1)(x-2)}. \]

Thus, we have:
\[ \frac{(x-3)(x-4) + (x-1)(x-4) + (x-1)(x-2)}{(x-1)(x-2)(x-3)(x-4)} = \frac{1}{6}. \]

### Step 2: Simplify the numerator
Now, we simplify the numerator:
\[ (x-3)(x-4) + (x-1)(x-4) + (x-1)(x-2). \]

Expanding each term:
1. \((x-3)(x-4) = x^2 - 7x + 12\),
2. \((x-1)(x-4) = x^2 - 5x + 4\),
3. \((x-1)(x-2) = x^2 - 3x + 2\).

Adding these together:
\[ (x^2 - 7x + 12) + (x^2 - 5x + 4) + (x^2 - 3x + 2) = 3x^2 - 15x + 18. \]

### Step 3: Set the equation
Now we set the equation:
\[ \frac{3x^2 - 15x + 18}{(x-1)(x-2)(x-3)(x-4)} = \frac{1}{6}. \]

Cross-multiplying gives:
\[ 6(3x^2 - 15x + 18) = (x-1)(x-2)(x-3)(x-4). \]

### Step 4: Expand the right-hand side
The right-hand side can be expanded as:
\[ (x-1)(x-4) = x^2 - 5x + 4, \]
\[ (x-2)(x-3) = x^2 - 5x + 6. \]

Now multiplying these two results:
\[ (x^2 - 5x + 4)(x^2 - 5x + 6). \]

Let \(y = x^2 - 5x\):
\[ (y + 4)(y + 6) = y^2 + 10y + 24. \]

Substituting back gives:
\[ (x^2 - 5x)^2 + 10(x^2 - 5x) + 24. \]

### Step 5: Rearranging the equation
Now we have:
\[ 18x^2 - 90x + 108 = (x^2 - 5x)^2 + 10(x^2 - 5x) + 24. \]

This is a quadratic equation in \(x\). Rearranging gives:
\[ 0 = (x^2 - 5x)^2 + 10(x^2 - 5x) + 24 - 18x^2 + 90x - 108. \]

### Step 6: Solve the quadratic equation
Now we can solve this quadratic equation using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}. \]

### Step 7: Find the roots
After solving, we find the roots to be \(x = 7\) and \(x = -2\).

### Final Answer:
Thus, the solutions for \(x\) are:
\[ x = 7 \quad \text{and} \quad x = -2. \]