Problem 106 - Algebra and Trigonometry

Question

perform the indicated operations. $$ \frac{1}{x^{n}-1}-\frac{1}{x^{n}+1}-\frac{1}{x^{2 n}-1} $$

Step-by-Step Solution

Answer

The simplified form of the given expression is \(\frac{1}{x^{2n}-1}\).

Step 1: Find the Least Common Denominator (LCD)

A fraction can only be simplified or subtracted if the denominators of all the fractions involved are the same. So the first step is to find the Least Common Denominator (LCD) for the fractions: \(x^n-1\), \(x^n+1\), and \(x^{2n}-1\). Through inspection, the LCD can be identified as \(x^{2n}-1\). This is because \(x^{2n}-1\) can be factored to \((x^n-1)(x^n+1)\), the denominators of the other two fractions.

Step 2: Rewrite the Fractions

Using the LCD, write each fraction with the denominator as \(x^{2n}-1\). Hence, the initial expression becomes:\(\frac{x^n+1-x^n+1-1}{x^{2n}-1} = \frac{1}{x^{2n}-1}\).

Step 3: Final simplification

The result from step 2 is the most simplified form of the given expression. Therefore, the final answer is \(\frac{1}{x^{2n}-1}\).

Key Concepts

Least Common Denominator (LCD)

Understanding how to find the Least Common Denominator (LCD) is crucial when working with algebraic fraction operations. The LCD is the smallest expression that each denominator can divide into without leaving a remainder. It's like finding a common ground where all fractions can be compared or combined.

Let's consider our example of combining these fractions: \( \frac{1}{x^{n}-1} \), \( \frac{1}{x^{n}+1} \), and \( \frac{1}{x^{2 n}-1} \). To perform any operation on these fractions, they must have the same denominator. By inspecting the given denominators, we determine that the LCD is \( x^{2n}-1 \), which is the product of the other two denominators \( x^n-1 \) and \( x^n+1 \).

Once the LCD is identified, it acts as a bridge allowing us to rewrite each fraction with a common denominator. This simplifies the process of adding, subtracting, or comparing fractions significantly. Being thorough in identifying the LCD can prevent errors and confusion deeper into solving algebraic problems.

Simplifying Fractions

Simplifying fractions is a process of reducing them to their most basic form, where the numerator and the denominator have no common factors other than 1. This makes the expression cleaner and often easier to work with. The simplification of fractions involves dividing both the top and bottom numbers by their greatest common factor until no further reduction is possible.

For instance, in the given problem: \( \frac{1}{x^{n}-1}-\frac{1}{x^{n}+1}-\frac{1}{x^{2 n}-1} \), after we find a common denominator, we combine the numerators over this common denominator and then simplify. In our example, the numerators combined to \( x^n+1-x^n+1-1 \), which simplified to 1. Hence, we ended with \( \frac{1}{x^{2n}-1} \).

It is important to always simplify fractions to their lowest terms, as this aids in comparison, further calculations, and provides a more elegant and precise result. Remember, the key to successful simplification is looking for common factors and reducing the fraction step by step until no further reduction is possible.

Polynomial Expressions

Polynomial expressions are algebraic expressions that involve variables raised to whole number exponents, and possibly constants, added or subtracted. They are the cornerstone for many areas of mathematics and are particularly prevalent in algebra.

In the context of the provided exercise, the denominators of the fractions are polynomial expressions. Recognizing the structure of polynomials can help in factoring and performing operations such as finding the LCD. For example, the expression \( x^{2n}-1 \) can be recognized as a difference of squares, a common polynomial identity, and can be factored into \( (x^n-1)(x^n+1) \).

Understanding the relationships and patterns within polynomial expressions allows us to manipulate and simplify algebraic fractions effectively. When operations involve polynomial denominators, factoring them can lead to significant simplifications, as we saw in converting the expression into a single fraction with a common denominator, thus easing the path to the solution.