Problem 46 - Elementary Algebra

Question

For the following problems, perform the divisions. $$ \frac{x^{3}-1}{x-1} $$

Step-by-Step Solution

Answer

Answer: The simplified expression is \(x^2 + x + 1\).

Step 1: Recall the Difference of Cubes Formula

The formula for the difference of cubes is given below: $$ a^3 - b^3 = (a - b)(a^2 + ab + b^2) $$ For our given expression, \(a = x\) and \(b = 1\). We will use this formula to factor the numerator.

Step 2: Factor the Numerator

Using the difference of cubes formula, we factor the numerator as follows: $$ x^3 - 1^3 = (x - 1)(x^2 + x\cdot1 + 1^2) = (x-1)(x^2 + x + 1) $$

Step 3: Replace the Numerator in the Original Expression

Replace the numerator with the factored expression we found in Step 2: $$ \frac{x^3-1}{x-1} = \frac{(x-1)(x^2+x+1)}{x-1} $$

Step 4: Cancel the Common Factors

Notice that we have a common factor of \((x-1)\) in both the numerator and denominator. Cancel this factor out: $$ \frac{(x-1)(x^2+x+1)}{x-1} = x^2+x+1 $$ Thus, the simplified expression after performing the division is: $$ x^2+x+1 $$

Key Concepts

Difference of Cubes

The difference of cubes is a special way of factoring a polynomial. It is specifically used when we have an expression like \(a^3 - b^3\). This technique allows us to break down the expression into two factors, which can make simplifying or solving much easier. In our problem, \(x^3 - 1\) resembles the form \(a^3 - b^3\) where \(a = x\) and \(b = 1\).
Let's explore the formula:

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

This formula is particularly useful because it transforms a higher-degree polynomial into a simpler multiplication of binomials and trinomials. Understanding how to apply this formula can provide a solid foundation in polynomial division and offers a shortcut when dealing with similarly structured problems.

Factoring

Factoring involves rewriting an expression as a product of its simpler factors. In the context of polynomial division, factoring is an essential step to simplify expressions. In our original exercise, we are tasked with the division of \(\frac{x^3-1}{x-1}\). The successful application of the difference of cubes allows us to factor \(x^3 - 1\) into \((x - 1)(x^2 + x + 1)\).
The process of factoring can be thought of as the reverse of expanding algebraic expressions. It seeks to answer, "What can I multiply together to get this?" When factoring, especially with polynomials, the goal is to express the polynomial as a product of its roots or simpler polynomial expressions. Once accomplished, these factors can then be used to simplify and solve equations more efficiently.

• Look for common factors.
• Apply special formulas (like the difference of cubes).
• Break down complex expressions into simpler components.

Simplifying Expressions

Simplifying expressions is the process of reducing them to their simplest or most compact form, without changing the value of the expression. In our exercise, after factoring, we have the expression \(\frac{(x-1)(x^2+x+1)}{x-1}\).
It is crucial to check for any common factors between the numerator and the denominator before simplifying. In this case, the common factor of \((x-1)\) can be canceled, resulting in the simplified expression \(x^2 + x + 1\).
Simplifying not only makes expressions easier to work with but also often reveals insights about the problem, such as roots or zero points.

• Identify and cancel common factors.
• Rewrite in the simplest form.
• Ensure that the value of the expression remains unchanged.