If a + b + c + d = 4 , the find the value of `(1)/(( 1 - a) ( 1 - b) (1 - c) )+ (1)/(( 1 - b) (1 - c) (1 - d)) + (1)/(( 1 - c) (1 - d) (1 - a)) + (1)/( ( 1 - d) (1 - a) (1 - b))`

Options

A. 0
B. 5
C. 1
D. 4

Answer

To solve the given problem, we need to evaluate the expression:

\[ \frac{1}{(1-a)(1-b)(1-c)} + \frac{1}{(1-b)(1-c)(1-d)} + \frac{1}{(1-c)(1-d)(1-a)} + \frac{1}{(1-d)(1-a)(1-b)} \]

given that \( a + b + c + d = 4 \).

### Step 1: Understand the relationship between a, b, c, and d
We know that \( a + b + c + d = 4 \). This means that if we express \( d \) in terms of \( a, b, \) and \( c \), we get:

\[ d = 4 - (a + b + c) \]

### Step 2: Substitute d in the expression
We can substitute \( d \) into the expression:

\[ \frac{1}{(1-a)(1-b)(1-c)} + \frac{1}{(1-b)(1-c)(1-(4-(a+b+c)))} + \frac{1}{(1-c)(1-(4-(a+b+c)))(1-a)} + \frac{1}{(1-(4-(a+b+c)))(1-a)(1-b)} \]

This simplifies to:

\[ \frac{1}{(1-a)(1-b)(1-c)} + \frac{1}{(1-b)(1-c)(1-(4-a-b-c))} + \frac{1}{(1-c)(1-(4-a-b-c))(1-a)} + \frac{1}{(1-(4-a-b-c))(1-a)(1-b)} \]

### Step 3: Simplify the terms
Notice that \( 1 - (4 - (a + b + c)) = a + b + c - 3 \). Therefore, we can rewrite the expression as:

\[ \frac{1}{(1-a)(1-b)(1-c)} + \frac{1}{(1-b)(1-c)(a+b+c-3)} + \frac{1}{(1-c)(a+b+c-3)(1-a)} + \frac{1}{(a+b+c-3)(1-a)(1-b)} \]

### Step 4: Find a common denominator
The common denominator for all these fractions will be:

\[ (1-a)(1-b)(1-c)(a+b+c-3) \]

### Step 5: Combine the fractions
Now we can combine the numerators over the common denominator. Each term will contribute to the numerator based on the missing factors:

\[ \frac{(a+b+c-3)(1-b)(1-c) + (1-a)(1-c)(1-b) + (1-a)(1-b)(1-c)}{(1-a)(1-b)(1-c)(a+b+c-3)} \]

### Step 6: Evaluate the numerator
We know that \( a + b + c = 4 - d \). Therefore, when we substitute \( d = 4 - (a + b + c) \), we find that the numerator simplifies down to zero because the terms cancel out.

### Step 7: Conclusion
Since the numerator is zero, the entire expression evaluates to:

\[ \frac{0}{(1-a)(1-b)(1-c)(a+b+c-3)} = 0 \]

Thus, the final answer is:

\[ \boxed{0} \]