Problem 66 - Elementary and Intermediate Algebra

Question

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

Step-by-Step Solution

Answer

The expression \(x^6-y^6\) factors as \((x-y)(x^2+xy+y^2)(x+y)(x^2-xy+y^2)\).

Step 1: Identify the expression as a difference of squares

Notice that the given expression is a difference of two terms: \(x^6 - y^6\). This expression fits the form \(a^2 - b^2\), where \(a = x^3\) and \(b = y^3\), because \((x^3)^2 = x^6\) and \((y^3)^2 = y^6\).

Step 2: Apply the difference of squares formula

Use the formula \(a^2 - b^2 = (a-b)(a+b)\) to factor the expression. Here, substitute \(a = x^3\) and \(b = y^3\) into the formula to factor \(x^6 - y^6\) as \((x^3-y^3)(x^3+y^3)\).

Step 3: Factor each resulting expression

The expression \(x^3 - y^3\) can be factored as a difference of cubes, \(x^3 - y^3 = (x-y)(x^2+xy+y^2)\). Similarly, \(x^3 + y^3\) can be factored as a sum of cubes, \(x^3 + y^3 = (x+y)(x^2-xy+y^2)\).

Step 4: Combine the factors

Substitute the factored forms of \(x^3 - y^3\) and \(x^3 + y^3\) back into the expression to get the complete factorization of \(x^6 - y^6\) as \( (x-y)(x^2+xy+y^2)(x+y)(x^2-xy+y^2)\).

Key Concepts

Difference of Squares

The concept of the difference of squares is a key factorization technique. It involves expressions where two squared terms are subtracted from one another. The general form is expressed as \(a^2 - b^2\). This can be factored into the product of two expressions: \((a-b)(a+b)\).

Using this formula allows us to simplify and solve many algebraic equations. In the original exercise, the expression \(x^6 - y^6\) is initially identified as a difference of squares. While it might not look like this at first glance, by recognizing that \((x^3)^2 = x^6\) and \((y^3)^2 = y^6\), we align it with the \(a^2 - b^2\) structure, letting us apply the formula effectively.

Applying this technique helps to break down complex expressions like \(x^6 - y^6\) into simpler, more manageable parts that can be factored further.

Difference of Cubes

Once an expression is reduced using the difference of squares, you might still encounter components that can be further simplified. This is where the difference of cubes comes in.

When we have an expression of the form \(a^3 - b^3\), it can be factored into \((a-b)(a^2+ab+b^2)\). In the original exercise, after using the difference of squares, we ended up with \((x^3-y^3)\) which can be factored as a difference of cubes. This additional step in the factorization process allows us to further simplify the expression.

The success of applying this formula depends on breaking down expressions into their cube components correctly, ensuring no mistakes are made while substituting values. For example,

• Identify \(a\) and \(b\) such that \(a^3 = x^3\) and \(b^3 = y^3\).
• Apply the formula to achieve a result: \((x-y)(x^2+xy+y^2)\).

Recognizing and applying the difference of cubes formula is crucial for complete factorization.

Sum of Cubes

Alongside the difference of cubes, the sum of cubes is another essential factorization method. It helps break down expressions of the form \(a^3 + b^3\). For this type, the formula used is \((a+b)(a^2-ab+b^2)\).

In our exercise involving \(x^3+y^3\), this approach further refines the expression.

• Similarly to the difference of cubes, identify the correct values \(a\) and \(b\) that align with the cube terms.
• Apply the sum of cubes formula: \((x+y)(x^2-xy+y^2)\).

By combining this with the previously found factors from the difference of cubes, the original expression \(x^6 - y^6\) achieves complete factorization. The systematic application of both the difference and sum of cubes, following the initial factorization by the difference of squares, enables a more straightforward and effective breakdown of complex algebraic expressions.