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Ch. R - Review of Basic Concepts
Lial - College Algebra 13th Edition
Lial13th EditionCollege AlgebraISBN: 9780136881063Not the one you use?Change textbook
All textbooksLial 13th EditionCh. R - Review of Basic ConceptsProblem 82
Chapter 1, Problem 82

Factor each polynomial. See Examples 5 and 6. 125-(4a-b)3

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Recognize that the expression \(125 - (4a - b)^3\) is a difference of cubes, since \(125 = 5^3\) and \((4a - b)^3\) is already a cube.
Recall the difference of cubes factoring formula: $x^3 - y^3 = (x - y)(x^2 + xy + y^2)$.
Identify \(x = 5\) and \(y = (4a - b)\) in the expression \(125 - (4a - b)^3\).
Apply the formula to factor the expression as: \((5 - (4a - b)) \left(5^2 + 5(4a - b) + (4a - b)^2\right)\).
Simplify the factors inside the parentheses by performing the arithmetic and expanding \((4a - b)^2\) to complete the factorization.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Difference of Cubes Formula

The difference of cubes formula states that a³ - b³ = (a - b)(a² + ab + b²). It is used to factor expressions where one cube is subtracted from another. Recognizing the structure allows you to rewrite the polynomial as a product of a binomial and a trinomial.
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Identifying Perfect Cubes

To apply the difference of cubes formula, you must identify terms that are perfect cubes. For example, 125 is 5³, and (4a - b)³ is the cube of the binomial (4a - b). Recognizing these helps in rewriting the expression in the form a³ - b³.
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Factoring Binomial Cubes

When factoring expressions like (4a - b)³, treat the entire binomial as a single term raised to the third power. This understanding is crucial for correctly applying the difference of cubes formula and expanding or factoring complex polynomial expressions.
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Related Practice
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