(a%2B1)(a2%2B1)(a4%2B1).jpeg "(a-1)(a+1)(a2+1)(a4+1)")
Exploring the Pattern: (a-1)(a+1)(a2+1)(a4+1)
This expression holds a fascinating pattern that leads to a surprisingly simple result. Let's delve into it:
Recognizing the Difference of Squares
Notice that the first two factors, (a-1) and (a+1), are in the form of a difference of squares:
- (a-1)(a+1) = a² - 1²
Similarly, the next two factors can be manipulated to fit the same pattern:
- (a² + 1)(a⁴ + 1) = (a²)² + 1²
Applying the Difference of Squares Pattern Repeatedly
Now, we can apply the difference of squares pattern again:
- a² - 1² = (a + 1)(a - 1)
- (a²)² + 1² = (a² + 1)(a² - 1)
Substituting these back into the original expression:
- (a-1)(a+1)(a² + 1)(a⁴ + 1) = (a + 1)(a - 1)(a² + 1)(a² - 1)
Simplifying the Expression
Continuing the process of applying the difference of squares pattern, we get:
- (a + 1)(a - 1)(a² + 1)(a² - 1) = (a² - 1²)(a⁴ - 1²) = (a² - 1)(a⁴ - 1)
Finally:
- (a² - 1)(a⁴ - 1) = a⁶ - 1
The Final Result
Therefore, the simplified form of the expression (a-1)(a+1)(a2+1)(a4+1) is a⁶ - 1. This pattern demonstrates the power of recognizing and applying algebraic identities to simplify complex expressions.