(a-c)-b(2a-b)-(a-b%2Bc)(a-b-c)%3D0.jpeg "(a+c)(a-c)-b(2a-b)-(a-b+c)(a-b-c)=0")
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.