(x-1)*(x^2+x+1)*(x^6+x^3+1)

Jasonbradley

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Jun 17, 2024

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The Curious Case of (x-1)(x^2+x+1)(x^6+x^3+1)

This seemingly complex expression holds some interesting mathematical properties, making it a fascinating subject for algebraic exploration. Let's break down its components and delve into its intriguing characteristics:

Factoring and the Difference of Cubes

The expression is a product of three distinct factors:

• (x-1): This is a simple linear factor.
• (x^2+x+1): This is a quadratic factor that cannot be further factored over real numbers.
• (x^6+x^3+1): This is a sixth-degree polynomial that can be factored using the "difference of cubes" pattern:
  • a^3 - b^3 = (a-b)(a^2 + ab + b^2)

Applying the difference of cubes pattern with a = x^2 and b = 1, we get:

(x^6 + x^3 + 1) = (x^2 - 1)(x^4 + x^2 + 1)

The first factor, (x^2 - 1), further simplifies to (x+1)(x-1), giving us:

(x^6 + x^3 + 1) = (x+1)(x-1)(x^4 + x^2 + 1)

The Full Factorization

Combining all the factors, the expression becomes:

(x-1)(x^2+x+1)(x^6+x^3+1) = (x-1)(x^2+x+1)(x+1)(x-1)(x^4 + x^2 + 1)

Simplifying:

(x-1)(x^2+x+1)(x^6+x^3+1) = (x-1)^2 (x+1)(x^2+x+1)(x^4 + x^2 + 1)

Key Observations

• Repeated Factor: The expression contains a repeated factor of (x-1) squared.
• Irreducible Factors: The factors (x^2+x+1) and (x^4 + x^2 + 1) are irreducible over real numbers. This means they cannot be further factored into expressions with real coefficients.
• Complex Roots: The irreducible factors have complex roots.

Conclusion

The seemingly complex expression (x-1)(x^2+x+1)(x^6+x^3+1) can be factored into a product of five factors, including a repeated factor and irreducible factors with complex roots. Understanding its factorization reveals its unique mathematical properties and highlights the importance of algebraic techniques for simplifying and analyzing complex expressions.