## Simplifying the Expression: (a+c)(a-c)-b(2a-b)-(a-b+c)(a-b-c)=0

This equation presents a challenge in its current form. To solve it, we can simplify the expression by expanding the products and combining like terms. Let's break it down step by step:

### Expanding the Products:

**(a+c)(a-c):**This is a difference of squares pattern, which simplifies to a² - c².**b(2a-b):**Distribute the 'b' to get 2ab - b².**(a-b+c)(a-b-c):**This requires careful expansion. We can use the FOIL method (First, Outer, Inner, Last) or the distributive property:**First:**a * a = a²**Outer:**a * (-b-c) = -ab - ac**Inner:**(-b+c) * a = -ab + ac**Last:**(-b+c) * (-b-c) = b² - c²- Combining terms: a² - 2ab + b² - c²

### Rewriting the Equation:

Now, our equation looks like this:

a² - c² - 2ab + b² - (a² - 2ab + b² - c²) = 0

### Simplifying and Solving:

Notice that many terms cancel out:

- a² cancels with -a²
- -2ab cancels with 2ab
- b² cancels with b²
- -c² cancels with c²

This leaves us with: **0 = 0**

### Conclusion

The equation (a+c)(a-c)-b(2a-b)-(a-b+c)(a-b-c)=0 simplifies to 0 = 0, indicating that the equation is true for **all** values of 'a', 'b', and 'c'. This means the equation is an **identity**, a statement that is always true.