Factorization Of 81a^6 - 100 A Step-by-Step Guide
In the realm of algebra, factorization plays a pivotal role in simplifying complex expressions and solving equations. Specifically, when we encounter expressions involving higher powers and constants, mastering factorization techniques becomes essential. Our focus in this article is to dissect the factorization of the expression 81a^6 - 100, a classic example that showcases the application of the difference of squares pattern. This pattern is a cornerstone in algebraic manipulations, and understanding its nuances is crucial for students and enthusiasts alike. We will embark on a step-by-step journey, exploring the fundamental concepts, applying relevant formulas, and ultimately arriving at the correct factorization. By the end of this discussion, you will not only grasp the solution to this particular problem but also gain a broader understanding of factorization strategies applicable to a wide range of algebraic expressions. This article aims to provide a comprehensive guide, ensuring clarity and fostering a deeper appreciation for the elegance of algebraic factorization.
Understanding the Difference of Squares
The difference of squares is a fundamental concept in algebra, which states that the difference between two perfect squares can be factored into the product of their sum and difference. Mathematically, this can be expressed as:
This formula is a cornerstone of factorization and is frequently used to simplify algebraic expressions. Recognizing the difference of squares pattern is the first step towards efficient factorization. In the context of our problem, 81a^6 - 100, we need to identify if both terms are perfect squares and if they are indeed separated by a subtraction sign. The term 81a^6 can be seen as the square of 9a^3, since (9a3)2 = 81a^6. Similarly, 100 is a perfect square, being the square of 10. Thus, we can rewrite the expression as (9a3)2 - (10)^2, which clearly fits the difference of squares pattern. The next step involves applying the formula directly, substituting 9a^3 for 'a' and 10 for 'b'. This initial transformation is crucial as it sets the stage for the subsequent steps in the factorization process. By understanding and recognizing this pattern, we lay a strong foundation for solving more complex problems in algebra. This foundational knowledge is not only crucial for academic success but also for various applications in engineering, physics, and other scientific disciplines where algebraic manipulations are frequently required.
Step-by-Step Factorization of 81a^6 - 100
To factorize 81a^6 - 100, we will meticulously follow the difference of squares pattern. As established earlier, the expression can be rewritten as (9a3)2 - (10)^2. This form directly corresponds to the difference of squares formula, a^2 - b^2 = (a - b)(a + b), where 'a' is 9a^3 and 'b' is 10. Applying the formula, we substitute these values into the factored form, resulting in (9a^3 - 10)(9a^3 + 10). This step is the heart of the factorization process, where the original expression is transformed into a product of two binomials. Upon closer inspection of these binomials, we need to ascertain if further factorization is possible. In this case, neither (9a^3 - 10) nor (9a^3 + 10) can be factored further using elementary algebraic techniques. This is because neither term fits any common factorization patterns such as the difference of cubes or sum of cubes, nor can they be simplified by extracting common factors. Therefore, the factorization process concludes here, and we have successfully expressed 81a^6 - 100 as the product of (9a^3 - 10) and (9a^3 + 10). This step-by-step approach ensures clarity and precision in the factorization process, highlighting the importance of recognizing patterns and applying formulas correctly. The resulting factors are the simplest forms achievable through algebraic manipulation, thus completing the factorization task.
Detailed Solution and Answer
Following the step-by-step factorization process, we arrive at the solution for the expression 81a^6 - 100. As we have established, the expression fits the difference of squares pattern and can be rewritten as (9a3)2 - (10)^2. Applying the difference of squares formula, a^2 - b^2 = (a - b)(a + b), where a = 9a^3 and b = 10, we obtain the factors (9a^3 - 10)(9a^3 + 10). This result corresponds to option C in the given choices. To verify the correctness of our factorization, we can multiply the two factors together and check if the product equals the original expression. Multiplying (9a^3 - 10) by (9a^3 + 10) yields:
(9a^3 - 10)(9a^3 + 10) = (9a3)(9a3) + (9a^3)(10) - (10)(9a^3) - (10)(10)
Simplifying this, we get:
81a^6 + 90a^3 - 90a^3 - 100 = 81a^6 - 100
Since the product of the factors matches the original expression, we can confidently conclude that our factorization is correct. Thus, the correct factorization of 81a^6 - 100 is indeed (9a^3 - 10)(9a^3 + 10), making option C the accurate answer. This detailed solution not only provides the answer but also reinforces the understanding of the factorization process and the verification method, ensuring a comprehensive grasp of the concept.
Why Other Options are Incorrect
To fully understand the solution, it's crucial to analyze why the other options are incorrect. This analysis reinforces our understanding of the correct method and helps prevent similar mistakes in the future. Let's examine each incorrect option:
-
Option A: (9a^2 - 10)(9a^3 + 10)
This option incorrectly identifies the square root of 81a^6. The square root of 81a^6 is 9a^3, not 9a^2. This misunderstanding leads to an incorrect application of the difference of squares formula. If we were to expand this product, we would not obtain the original expression, 81a^6 - 100. The presence of the a^2 term in the first factor indicates a flaw in the initial identification of the perfect square.
-
Option B: (9a^2 - 10)(9a^3 - 10)
Similar to option A, this option also misidentifies the square root of 81a^6 as 9a^2. Additionally, this option presents both factors with a subtraction sign, which is inconsistent with the difference of squares pattern. The difference of squares formula requires one factor to be the sum and the other to be the difference. Expanding this product would result in terms that do not cancel out to give the original expression, thus confirming its incorrectness.
-
Option D: (9a^3 - 10)(9a^3 - 10)
This option correctly identifies the square root of 81a^6 as 9a^3 but incorrectly applies the difference of squares formula. It presents both factors as a difference, which is akin to squaring the binomial (9a^3 - 10). Expanding this product would yield (9a^3 - 10)^2 = 81a^6 - 180a^3 + 100, which is clearly not equal to the original expression 81a^6 - 100. This option overlooks the crucial aspect of having both a sum and a difference in the factors when applying the difference of squares.
By dissecting these incorrect options, we gain a deeper appreciation for the correct application of the difference of squares formula and the importance of accurately identifying perfect squares. This comparative analysis enhances our problem-solving skills and reinforces the understanding of algebraic factorization.
Conclusion
In summary, the correct factorization of the expression 81a^6 - 100 is (9a^3 - 10)(9a^3 + 10), which corresponds to option C. This factorization is achieved by recognizing and applying the difference of squares pattern, a fundamental concept in algebra. Throughout this discussion, we have emphasized the importance of accurately identifying perfect squares and applying the formula a^2 - b^2 = (a - b)(a + b) correctly. We meticulously walked through the step-by-step process, ensuring clarity and precision at each stage. Furthermore, we analyzed why the other options are incorrect, reinforcing the correct application of the difference of squares formula and highlighting common pitfalls to avoid. Mastering factorization techniques like this is crucial for success in algebra and beyond. These skills are not only essential for solving equations and simplifying expressions but also form the foundation for more advanced mathematical concepts. By understanding the underlying principles and practicing regularly, you can confidently tackle a wide range of factorization problems. This article serves as a comprehensive guide, equipping you with the knowledge and understanding necessary to excel in algebraic manipulations and problem-solving.
