If a, b and c are distinct positive numbers, then the expression `(a + b - c)(b+ c- a)(c+ a -b)- abc` is:

Options

A. positive
B. negative
C. non-positive
D. non-negative

Explanation

To solve the problem, we need to analyze the expression \((a + b - c)(b + c - a)(c + a - b) - abc\) under the condition that \(a\), \(b\), and \(c\) are distinct positive numbers.

### Step-by-Step Solution:

1. **Understanding the Terms**:
We have three terms:
- \(x_1 = a + b - c\)
- \(x_2 = b + c - a\)
- \(x_3 = c + a - b\)

Each of these terms represents a difference between the sum of two numbers and a third number.

2. **Analyzing the Positivity of Each Term**:
Since \(a\), \(b\), and \(c\) are distinct positive numbers, we can analyze the signs of these terms:
- \(x_1 = a + b - c\) can be positive if \(a + b > c\).
- \(x_2 = b + c - a\) can be positive if \(b + c > a\).
- \(x_3 = c + a - b\) can be positive if \(c + a > b\).

By the triangle inequality, all three conditions hold true for distinct positive \(a\), \(b\), and \(c\). Thus, \(x_1\), \(x_2\), and \(x_3\) are all positive.

3. **Applying the AM-GM Inequality**:
We can apply the Arithmetic Mean-Geometric Mean (AM-GM) inequality:
\[ \frac{x_1 + x_2}{2} \geq \sqrt{x_1 x_2} \]
\[ \frac{x_2 + x_3}{2} \geq \sqrt{x_2 x_3} \]
\[ \frac{x_3 + x_1}{2} \geq \sqrt{x_3 x_1} \]

Since \(x_1\), \(x_2\), and \(x_3\) are positive, we can derive inequalities that relate the products of these terms.

4. **Multiplying the Inequalities**:
By multiplying the inequalities derived from AM-GM, we can establish a relationship:
\[ abc < (a + b - c)(b + c - a)(c + a - b) \]

5. **Rearranging the Expression**:
From the above inequality, we can rearrange it to get:
\[ (a + b - c)(b + c - a)(c + a - b) - abc > 0 \]

6. **Conclusion**:
Therefore, we conclude that the expression \((a + b - c)(b + c - a)(c + a - b) - abc\) is positive.

### Final Answer:
The expression \((a + b - c)(b + c - a)(c + a - b) - abc\) is **positive**.