Problem 88 - Introductory Algebra for College Students

Question

Factor using the formula for the sum or difference of two cubes. $$x^{3} y^{3}-27$$

Step-by-Step Solution

Answer

The factorization of \(x^{3} y^{3} - 27\) is \((xy - 3)(x^{2}y^{2} + 3xy + 9)\).

Step 1: Rewrite the expression as the difference of two cubes

Firstly, we can rewrite \(x^{3} y^{3} - 27\) as \({(xy)}^3 - 3^3\). In this case, \(a = xy\) and \(b = 3\).

Step 2: Apply the formula for the difference of cubes

We apply the formula for difference of cubes which is \(a^{3} - b^{3} = (a - b)(a^{2} + ab + b^{2})\). By doing this, we substitute \(a\) with \(xy\) and \(b\) with \(3\), we get \((xy - 3)(x^{2}y^{2} + 3xy + 9)\).

Step 3: Check the result

We can check our result by expanding our result to see whether it equals to our original expression, which is \(x^{3} y^{3} - 27\).

Key Concepts

Difference of Cubes

When we talk about the **difference of cubes**, we are discussing a specific way to handle a polynomial that fits the form of two terms, each raised to a power of three, separated by a subtraction sign. The general formula to factor a difference of cubes, such as \(a^3 - b^3\), is:

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

This formula provides a way to break down the expression into simpler components. In the exercise, the expression \(x^3 y^3 - 27\) needs to be recognized as a difference of cubes.

• The term \((xy)^3\) can be identified as one of the cubes.
• The number \(27\) is a perfect cube as it equals \(3^3\).

Simply substitute \(a = xy\) and \(b = 3\) into the formula and apply it accordingly. This step turns our polynomial into the factored form: \((xy - 3)(x^2y^2 + 3xy + 9)\). Whenever you see an equation structured like this with cubes and subtraction, remember the difference of cubes formula. It’s like having a handy toolkit for breaking down complex expressions.

Algebraic Expressions

**Algebraic expressions** are combinations of numbers, variables, and operations. They're like the building blocks of all algebraic calculations. In the given expression \(x^3 y^3 - 27\), you're working with:

• **Variables:** These are symbols that represent quantities, typically represented by letters like \(x\) and \(y\). They can change and are not fixed values.
• **Constants:** These are fixed numbers in an expression. Here \(27\) is a constant.
• **Operations:** These include addition, subtraction, multiplication, etc., which are used to combine the numbers and variables.

Identifying how to manipulate algebraic expressions is crucial. In this exercise, recognizing that \(x^3 y^3\) is a product of variables raised to a power is key to understanding it as a cube. This insight allows you to apply specific factoring strategies. By understanding the components and structure of algebraic expressions, you can better maneuver through various operations and transformations, such as factoring.

Polynomial Factoring

**Polynomial factoring** is a technique used to express polynomials as a product of simpler polynomials. It’s essential for simplifying expressions and solving equations. The exercise demonstrates polynomial factoring through the expression \((xy)^3 - 3^3\). Here's a bit of insight into the process:

• **Identify Each Term:** Notice whether the polynomial is made up of terms with variable powers, perfect squares, cubes, etc. Here you have cubes which lead you to consider the difference of cubes formula.
• **Apply Factoring Formulas:** Utilize known formulas like the difference of cubes to break down the polynomial into a product of factors. This makes it easier to handle.
• **Check Your Work:** Always expand the factored form back out to ensure it matches the original expression. This step validates your process and confirms the accuracy of the factorization.

In the given solution, factoring \((xy)^3 - 3^3\) using the difference of cubes formula beautifully collapses this complex polynomial into two simpler binomials \((xy - 3)(x^2y^2 + 3xy + 9)\). Understanding polynomial factoring not only aids in simplifying expressions but also reveals the relationships within expressions, clearing the path to solving them.