Problem 79 - Intermediate Algebra

Question

Perform the operations and simplify the result when possible. $$\frac{a^{2}+a b}{a^{3}-b^{3}}-\frac{b^{2}}{b^{3}-a^{3}}$$

Step-by-Step Solution

Answer

The simplified form is \(\frac{a^2 + ab + b^2}{a^3 - b^3}\).

Step 1: Identify the structure of the expression

The expression to simplify is \[\frac{a^{2}+a b}{a^{3}-b^{3}}-\frac{b^{2}}{b^{3}-a^{3}}.\]Recognize that the denominators, \(a^{3}-b^{3}\) and \(b^{3}-a^{3}\), are differences of cubes, which are related by a negative factor.

Step 2: Factor the differences of cubes

The formula for the difference of cubes is given by \(x^3 - y^3 = (x-y)(x^2 + xy + y^2)\). Apply this formula to factorize both denominators: \[a^3-b^3 = (a-b)(a^2+ab+b^2) \b^3-a^3 = -(a-b)(a^2+ab+b^2).\]Substitute these factored forms into the expression.

Step 3: Rewrite the expression using factored denominators

Substitute the factorized forms into the original expression:\[\frac{a^2+ab}{(a-b)(a^2+ab+b^2)} - \frac{b^2}{-(a-b)(a^2+ab+b^2)}.\]Note that the denominator of the second fraction is \(-(a-b)(a^2+ab+b^2)\).

Step 4: Combine the fractions

Both fractions have a common base denominator of \((a-b)(a^2+ab+b^2)\), but with the negative sign on the second term. Simplify the negative sign by rewriting the expression as:\[\frac{a^2 + ab}{(a-b)(a^2+ab+b^2)} + \frac{b^2}{(a-b)(a^2+ab+b^2)}.\]

Step 5: Add the numerators

Combine the numerators over the common denominator:\[\frac{(a^2 + ab) + b^2}{(a-b)(a^2+ab+b^2)} = \frac{a^2 + ab + b^2}{(a-b)(a^2+ab+b^2)}.\]The expression can be further simplified to:\[\frac{a^2 + ab + b^2}{a^3 - b^3}.\]

Step 6: Final Step: Verify and finalize the expression

Check each previous step for mathematical accuracy and ensure no further simplifications are possible. The resulting expression:\[\frac{a^2 + ab + b^2}{a^3 - b^3}\]is simplified as far as possible, as the numerator is equivalent to a part of the factorized form of the denominator's original structure.

Key Concepts

Difference of Cubes

To simplify expressions involving cube terms, it's crucial to recognize the "Difference of Cubes". This concept involves algebraic expressions where a cube is subtracted from another. The general form is given by \(x^3 - y^3\), and it can be factored into \((x-y)(x^2 + xy + y^2)\). This factoring is particularly useful because it simplifies complex denominators or numerators in fractions.

In the exercise, both denominators, \(a^3-b^3\) and \(b^3-a^3\), are classic examples of differences of cubes. Notice that the formula remains the same, while just swapping \(a\) and \(b\) offers the factored forms needed to combine or rearrange expressions efficiently.

• Always look for cube terms and attempt to factor them using the difference of cubes formula.
• The formula simplifies expressions significantly, allowing for easier arithmetic manipulation.

Remember, recognizing such patterns is a skill that improves algebraic problem-solving efficiency.

Common Denominator

Finding a common denominator is a key step in adding or subtracting fractions with different denominators. In essence, it allows us to combine fractions into a single fraction, simplifying the entire process.

In the exercise, once the denominators are factored, notice that \((a-b)(a^2 + ab + b^2)\) becomes the common denominator for both terms. The adjustment from \(-(a-b)(a^2+ab+b^2)\) to \((a-b)(a^2+ab+b^2)\) just requires a minor algebraic sign change.

• Having a common denominator streamlines solving and simplifying expressions involving addition or subtraction of fractions.
• This technique is foundational for complex algebraic manipulations.

Grasping this concept will make handling seemingly intricate expressions easier and less prone to errors.

Factoring Expressions

Factoring is the method of breaking down an expression into simpler 'factors' that, when multiplied together, yield the original expression. It's like reverse engineering a product to understand its components.

The exercise demonstrates factoring by taking the "Difference of Cubes" formula and applying it to rewrite the denominators. This reveals common factors and simplifies similar algebraic problems.

• Factoring is crucial because it turns complex expressions into more manageable forms.
• In algebra, factored forms often provide insights necessary for further simplification or problem-solving.

When faced with an algebraic expression, see if there's an opportunity to factor out common elements or structures. This technique aids not just simplification, but also in solving equations and proving identities.