Problem 55 - Algebra for College Students

Question

Use the sum-of-two-cubes or the difference-of-two-cubes pattern to factor each of the following. $$x^{6}-y^{6}$$

Step-by-Step Solution

Answer

Factored form: \((x-y)(x+y)(x^4 + x^2y^2 + y^4)\).

Step 1: Identify the Pattern

Observe the expression given: \(x^6 - y^6\). Notice that it is a difference of two cubes, since each term is a cube: \((x^2)^3 - (y^2)^3\). This hints at using the difference-of-two-cubes formula, which is \(a^3 - b^3 = (a - b)(a^2 + ab + b^2)\).

Step 2: Rewrite the Expression

Express \(x^6 - y^6\) as a difference of cubes: let \(a = x^2\) and \(b = y^2\), so \(x^6 - y^6 = (x^2)^3 - (y^2)^3\).

Step 3: Apply the Difference-of-Two-Cubes Formula

Using the formula \(a^3 - b^3 = (a - b)(a^2 + ab + b^2)\), where \(a = x^2\) and \(b = y^2\), factor the expression to get \((x^2 - y^2)((x^2)^2 + (x^2)(y^2) + (y^2)^2)\).

Step 4: Simplify the Factored Expression

Simplify the terms in the second factor: \((x^2)^2 + (x^2)(y^2) + (y^2)^2 = x^4 + x^2y^2 + y^4\), leading to the expression \((x^2 - y^2)(x^4 + x^2y^2 + y^4)\).

Step 5: Further Factor the First Term

Recognize the first factor \(x^2 - y^2\) as a difference of squares, which can be factored using \(a^2 - b^2 = (a-b)(a+b)\). Thus, \(x^2 - y^2 = (x-y)(x+y)\).

Step 6: Final Factorization

Combine everything to write the fully factored expression: \((x-y)(x+y)(x^4 + x^2y^2 + y^4)\).

Key Concepts

Difference of Cubes

When working with polynomial expressions, one common formula that we utilize is the difference of cubes. This formula helps in breaking down or factoring polynomials that are a difference between two cubed terms.

The difference of cubes is represented by the formula \(a^3 - b^3 = (a-b)(a^2 + ab + b^2)\). Here, \(a\) and \(b\) are the terms being cubed. A key point to remember is that this formula results in two distinct components:

• The first part, \((a-b)\), is a linear binomial.
• The second part, \((a^2 + ab + b^2)\), is a trinomial.

In the exercise, the expression \(x^6 - y^6\) was rewritten as \((x^2)^3 - (y^2)^3\) allowing for the difference of cubes method to be applied. By setting \(a = x^2\) and \(b = y^2\), we factor it as \((x^2 - y^2)(x^4 + x^2y^2 + y^4)\).

Understanding this concept is crucial as it facilitates the simplification of polynomial expressions, making them easier to work with. Factorization using the difference of cubes is often a step in solving more complex algebraic problems.

Difference of Squares

Another essential concept in algebra is the difference of squares. This offers a straightforward way to factor expressions that fit the mold \(a^2 - b^2\).

The difference of squares is expressed by the formula \(a^2 - b^2 = (a-b)(a+b)\). It works because the middle terms of the expanded product cancel each other out, leaving a simple difference of squares.

In resolving \(x^2 - y^2\) in the original problem, we apply this rule, knowing \(a = x\) and \(b = y\). Thus, it factors into \((x-y)(x+y)\). This step simplifies the expression further, breaking it down into basic binomials. This type of factorization is incredibly useful due to its simplicity and effectiveness in breaking down and simplifying polynomial expressions.

• Ensures easier manipulation of algebraic expressions.
• Frequently used in conjunction with other factorization techniques.
• Powerful when solving equations set to zero brackets to find roots.

Algebraic Expressions

Understanding algebraic expressions is fundamental to mastering mathematics topics like factorization. An algebraic expression is a combination of numbers, variables, and operators (such as plus and minus signs).

They form the basis of equations and formulas used throughout various branches of science and engineering. In algebra, we often examine these expressions to simplify or manipulate them using different methods and formulas.

Consider the initial expression \(x^6 - y^6\) from the task. This is a combination of powers of \(x\) and \(y\) and represents an algebraic expression that can be broken down using factorization techniques, such as the difference of cubes and squares. Recognizing patterns in algebraic expressions allows us to apply appropriate formulas to simplify or rearrange them.

• Key to solving equations efficiently.
• Enables understanding of the behavior of functions.
• Crucial for solving real-world problems analytically.