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Chapter 6: Problem 268

Factor completely. $$ y^{6}+1 $$

Short Answer

Expert verified
Factor completely: \( y^6 + 1 = (y^2 + 1)(y^4 - y^2 + 1) \).

Step by step solution

01

Recognize the Pattern

Identify that the expression is a sum of two cubes. Recall that \( a^3 + b^3 = (a + b)(a^2 - ab + b^2) \). Here, \( y^6 = (y^2)^3 \) and 1 = 1^3.
02

Apply the Sum of Cubes Formula

Use the sum of cubes formula with \( a = y^2 \) and \( b = 1 \). Therefore, \( y^6 + 1 = (y^2 + 1)((y^2)^2 - (y^2)(1) + 1^2) = (y^2 + 1)(y^4 - y^2 + 1) \).
03

Verify Each Factor

Confirm that \( y^2 + 1 \) and \( y^4 - y^2 + 1 \) are each completely factored. Neither expression can be factored further with real numbers.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Sum of Cubes, Polynomial Factorization, and Intermediate Algebra Problems
Factoring is one of the cornerstone skills in algebra. It's used to simplify expressions and solve algebraic equations. Here, we'll discuss core concepts related to the exercise of factoring the polynomial expression \(y^{6}+1\). Understanding these principles will make solving such problems easier and more intuitive.

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Most popular questions from this chapter

In the following exercises, foctor using substitution. $$ (5 y-1)^{2}-3(5 y-1)-18 $$

In the following exercises, factor using the 'ac' method. $$ 6 p^{2}+p-22 $$

In the following exercises, factor each expression using any method. $$ (x+3)^{2}-9(x+3)-36 $$

In the following exercises, factor each trinomial of the form \(x^{2}+b x+c .\) $$ q^{2}-13 q+36 $$

In the following exercises, factor each expression using any method. $$ x^{4}-4 x^{2}-12 $$
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