In Exercises 1–68, factor completely, or state that the polynomial is prime. x⁸ − y⁸

Verified step by step guidance

1. Recognize that the expression $x^8-y^8$ is a difference of two powers.
2. Recall the formula for factoring a difference of squares: $a^2-b^2=(a-b)(a+b)$.
3. Notice that $x^8-y^8$ can be rewritten as $(x^4{)}^2-(y^4{)}^2$, which is a difference of squares.
4. Apply the difference of squares formula: $(x^4-y^4)(x^4+y^4)$.
5. Factor $x^4-y^4$ further using the difference of squares: $(x^2-y^2)(x^2+y^2)$.

Key Concepts

Difference of Squares

The difference of squares is a fundamental algebraic identity stating that a² - b² can be factored into (a - b)(a + b). This concept is crucial for factoring polynomials that can be expressed as the difference between two perfect squares, which is applicable in the given polynomial x⁸ - y⁸.

Factoring Polynomials

Factoring polynomials involves rewriting a polynomial as a product of simpler polynomials or factors. Understanding how to identify common factors, apply special identities, and break down higher-degree polynomials is essential for completely factoring expressions like x⁸ - y⁸.

Prime Polynomials

A prime polynomial is one that cannot be factored into the product of two non-constant polynomials with real coefficients. Recognizing when a polynomial is prime is important in algebra, as it determines whether further factorization is possible, which is relevant when analyzing the polynomial x⁸ - y⁸.