(a%2Bb)%2B(b-c)(b%2Bc)%2B(c-a)(c%2Ba)%3D0.jpeg "(a-b)(a+b)+(b-c)(b+c)+(c-a)(c+a)=0")
The Elegance of (a-b)(a+b)+(b-c)(b+c)+(c-a)(c+a)=0
This equation may seem complex at first glance, but it hides a simple and beautiful mathematical truth. Let's break it down:
Recognizing the Pattern
The equation consists of three terms, each in the form of (x - y)(x + y). This pattern is crucial because it represents the difference of squares factorization:
(x - y)(x + y) = x² - y²
Applying the Difference of Squares
Applying this to our original equation, we get:
- (a - b)(a + b) = a² - b²
- (b - c)(b + c) = b² - c²
- (c - a)(c + a) = c² - a²
Substituting these back into the original equation:
(a² - b²) + (b² - c²) + (c² - a²) = 0
Simplifying the Equation
Notice that all the squared terms cancel each other out:
- a² - a² = 0
- b² - b² = 0
- c² - c² = 0
This leaves us with:
0 + 0 + 0 = 0
Conclusion
The equation (a-b)(a+b)+(b-c)(b+c)+(c-a)(c+a)=0 holds true for any values of a, b, and c. This is because the difference of squares pattern allows for elegant simplification, ultimately resulting in a trivial equality. This equation demonstrates the power of recognizing patterns in mathematics and how it can lead to beautiful and insightful results.