Factoring the Difference of n-th Powers
Разность и сумма n-х степеней
From familiar formulas to a powerful general pattern
§1. Review of Known Formulas
You already know two important factoring formulas. Let's write them side by side and look for a pattern:
Difference of Squares (n = 2)
a² − b² = (a − b)(a + b)
Second factor has 2 terms
Difference of Cubes (n = 3)
a³ − b³ = (a − b)(a² + ab + b²)
Second factor has 3 terms
🔍 Key Observation: In both cases, the first factor is (a − b). The second factor has as many terms as the exponent n. Can we generalize this to any power n?
§2. Building the Pattern: n = 4
Let's factor a⁴ − b⁴. There are two approaches:
Method 1: Repeated Difference of Squares
a⁴ − b⁴ = (a²)² − (b²)²
= (a² − b²)(a² + b²)
= (a − b)(a + b)(a² + b²)
Method 2: General Pattern
a⁴ − b⁴ = (a − b)(a³ + a²b + ab² + b³)
Second factor has 4 terms
Powers of a: 3, 2, 1, 0
Powers of b: 0, 1, 2, 3
✅ Both methods are correct! Method 2 follows the emerging pattern. Note that (a + b)(a² + b²) = a³ + a²b + ab² + b³, so the results agree.