Problem 65 - Elementary and Intermediate Algebra

Question

Factor each expression completely. Factor a difference of two squares first. \(x^{6}-1\)

Step-by-Step Solution

Answer

The expression \(x^6 - 1\) factors to \((x-1)(x^2 + x + 1)(x+1)(x^2 - x + 1)\).

Step 1: Identify the Difference of Squares

First, notice that the expression is in the form of a difference of squares: \(x^6 - 1^2\). This indicates that it can be factored using the formula for a difference of squares, which is \(a^2 - b^2 = (a-b)(a+b)\). Here, \(a = x^3\) and \(b = 1\).

Step 2: Apply the Difference of Squares Formula

Apply the difference of squares formula to factor the expression: \(x^6 - 1 = (x^3 - 1)(x^3 + 1)\). Now the expression has been factored into two terms, each of which is a cubic polynomial that can be further factored.

Step 3: Factor the Cubic Polynomial \(x^3 - 1 \)

The polynomial \(x^3 - 1\) is a difference of cubes, which can be factored using \(a^3 - b^3 = (a-b)(a^2 + ab + b^2)\). In this case, \(a = x\) and \(b = 1\). Applying the formula results in \(x^3 - 1 = (x-1)(x^2 + x + 1)\).

Step 4: Factor the Cubic Polynomial \(x^3 + 1 \)

The polynomial \(x^3 + 1\) is a sum of cubes, which can be factored using \(a^3 + b^3 = (a+b)(a^2 - ab + b^2)\). Here, \(a = x\) and \(b = 1\). This gives us \(x^3 + 1 = (x+1)(x^2 - x + 1)\).

Step 5: Combine the Factored Parts

Combine all the factored parts from the previous steps to achieve the completely factored form of the original expression: \(x^6 - 1 = (x-1)(x^2 + x + 1)(x+1)(x^2 - x + 1)\). Each part is now fully expanded and cannot be factored further.

Key Concepts

Difference of Squares

The difference of squares is a fundamental algebraic concept that is often used to simplify or factor expressions. When you see an expression like \(a^2 - b^2\), that's a clue it's in the difference of squares form. To factor it, you use the formula:

• \(a^2 - b^2 = (a-b)(a+b)\)

This formula is incredibly powerful because it allows you to break down complex expressions into simpler parts. For example, in the expression \(x^6 - 1\), you can recognize it as a difference of squares because \(x^6\) is \((x^3)^2\) and 1 is \(1^2\). Thus, applying the formula gives:

• \(x^6 - 1^2 = (x^3 - 1)(x^3 + 1)\)

By understanding this concept, you can quickly see how to simplify and approach the problem more effectively, setting you up for the next steps of factoring.

Difference of Cubes

The difference of cubes is another expression that can be factored using a specific formula. When an expression is in the form \(a^3 - b^3\), it can be factored using:

• \(a^3 - b^3 = (a-b)(a^2 + ab + b^2)\)

This formula allows you to break down a difference of cubes into a product of a linear term and a quadratic term. In the original exercise, the difference \(x^3 - 1\) fits this pattern where \(a = x\) and \(b = 1\). Using the formula, it becomes:

• \(x^3 - 1 = (x-1)(x^2 + x + 1)\)

It's essential to remember this formula because it simplifies the factoring process of higher degree polynomials into manageable components.

Sum of Cubes

The sum of cubes, like the difference of cubes, also has a specific formula used for factoring. An expression written in the form \(a^3 + b^3\) can be conveniently factored using:

• \(a^3 + b^3 = (a+b)(a^2 - ab + b^2)\)

This formula breaks down the sum of cubes into a product of a linear term and a quadratic factor. In our step-by-step solution, \(x^3 + 1\) is identified as a sum of cubes with \(a = x\) and \(b = 1\). By applying the formula, you get:

• \(x^3 + 1 = (x+1)(x^2 - x + 1)\)

Understanding how to factor sums of cubes is very useful for handling cubic expressions that initially seem more complex. With this know-how, you can simplify and solve equations more efficiently.