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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
Chapter 1, Problem 78

Factor each polynomial. See Examples 5 and 6. 27z9+64y1227z^9+64y^{12}

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1
Recognize that the polynomial \$27z^9 + 64y^{12}\( is a sum of two cubes because \)27z^9 = (3z^3)^3\( and \)64y^{12} = (4y^4)^3$.
Recall the sum of cubes factoring formula: \(a^3 + b^3 = (a + b)(a^2 - ab + b^2)\).
Identify \(a = 3z^3\) and \(b = 4y^4\) in the expression \$27z^9 + 64y^{12}$.
Apply the formula by substituting \(a\) and \(b\): write the factorization as \((3z^3 + 4y^4)((3z^3)^2 - (3z^3)(4y^4) + (4y^4)^2)\).
Simplify each term inside the second factor: \((3z^3)^2 = 9z^6\), \((3z^3)(4y^4) = 12z^3y^4\), and \((4y^4)^2 = 16y^8\), so the full factorization is \((3z^3 + 4y^4)(9z^6 - 12z^3y^4 + 16y^8)\).

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Key Concepts

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

Sum of Cubes Formula

The sum of cubes formula states that a³ + b³ = (a + b)(a² - ab + b²). It is used to factor expressions where two terms are perfect cubes added together. Recognizing the terms as cubes allows you to apply this formula to factor the polynomial completely.
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To factor using special formulas, you must identify if terms are perfect powers, such as perfect squares or cubes. For example, 27z⁹ is (3z³)³ and 64y¹² is (4y⁴)³. Recognizing these helps in rewriting the polynomial in a form suitable for applying the sum of cubes formula.
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