Problem 78 - Algebra for College Students

Question

Perform the indicated operations. $$\frac{x^{3}+1}{x^{2}-1} \cdot \frac{3 x-3}{x^{3}-x^{2}+x}$$

Step-by-Step Solution

Answer

\frac{3}{x}

Step 1: Factor all expressions

First, factor each part of the given expressions:- \require{x^3 + 1} can be factored using the sum of cubes:\[x^3 + 1 = (x + 1)(x^2 - x + 1)\]- \require{x^2 - 1} is a difference of squares:\[x^2 - 1 = (x - 1)(x + 1)\]- \require{3x - 3} is factorable:\[3x - 3 = 3(x - 1)\]- \require{x^3 - x^2 + x} can be factored by taking an x out:\[x^3 - x^2 + x = x(x^2 - x + 1)\]

Step 2: Substitute factored forms

Next, substitute these factored forms back into the original expression:\[\frac{(x + 1)(x^2 - x + 1)}{(x - 1)(x + 1)} \times \frac{3(x - 1)}{x(x^2 - x + 1)}\]

Step 3: Simplify the expression

Cancel out common factors in the numerator and denominator:- The \require{x + 1} terms cancel out- The \require{x^2 - x + 1} terms also cancel out- The \require{x - 1} terms cancel outThe simplified expression is: \[\frac{3}{x}\]

Step 4: Result

After simplifying, the result of the operation is:\[\frac{3}{x}\]

Key Concepts

Algebraic Operations

Algebraic operations include addition, subtraction, multiplication, and division of algebraic expressions. In this problem, we are dealing with multiplication and division.

When performing these operations, it's essential to simplify as much as possible. Simplifying helps in reducing complex expressions to their simplest form.

Consider the expressions given: \(\frac{x^3+1}{x^2-1} \times \frac{3x-3}{x^3-x^2+x}\). Our goal is to break these down using algebraic operations to make the expressions easier to handle.

For multiplication, \(\frac{a}{b} \times \frac{c}{d}\) where a, b, c, and d are algebraic expressions, we multiply the numerators together and the denominators together: \(\frac{a \times c}{b \times d}\). Simplifying before multiplying can significantly reduce the complexity of the process.

Similarly, division of expressions can be approached by multiplying by the reciprocal. Keep this in mind as it will come in handy when simplifying rational expressions and factoring polynomials.

Factoring Polynomials

Factoring is the process of breaking down polynomials into products of simpler polynomials. This is a crucial step in solving many algebraic problems.

In our problem, we need to factor four expressions:

• \[x^3 + 1\] (sum of cubes):\(x^3 + 1 = (x + 1)(x^2 - x + 1)\)
• \[x^2 - 1\] (difference of squares):\(x^2 - 1 = (x - 1)(x + 1)\)
• \[3x - 3\] (common factor):\(3x - 3 = 3(x - 1)\)
• \[x^3 - x^2 + x\] (common factor):\(x^3 - x^2 + x = x(x^2 - x + 1)\)

Factoring simplifies our problem by turning complicated expressions into products of linear or quadratic factors that are easier to handle.

Once we have factored all parts of the expression, we can substitute the factored forms back into the original equation. This gives us a new, more manageable expression to work with.

Simplifying Rational Expressions

Rational expressions are fractions where the numerator and the denominator are polynomials. Simplifying these expressions makes solving them much easier.

After substituting the factored forms back into the expression, we have:
\[\frac{(x + 1)(x^2 - x + 1)}{(x - 1)(x + 1)} \times \frac{3(x - 1)}{x(x^2 - x + 1)}\]

We then look for common factors in the numerator and the denominator:

• \(x + 1\) terms cancel out
• \(x^2 - x + 1\) terms cancel out
• \(x - 1\) terms cancel out

After canceling out these common factors, we are left with:
\[\frac{3}{x}\]

This simplified form \(\frac{3}{x}\) makes it evident that there's no further reduction possible.

Always remember to identify and cancel common factors to simplify the expression. This step is crucial as it transforms complex rational expressions into much simpler forms.