Factor 216p⁶ + 1: A Step-by-Step Guide

by Henrik Larsen 39 views

Hey guys! Today, we're diving deep into the world of factoring, and we're going to tackle a particularly interesting expression: 216p⁶ + 1. Factoring might seem daunting at first, but trust me, once you get the hang of it, it's like solving a puzzle – super satisfying! We'll break down the steps, explain the concepts, and by the end of this article, you'll be a factoring whiz. So, let's get started, shall we?

Understanding the Sum of Cubes

Before we jump into factoring 216p⁶ + 1, let's talk about a crucial concept: the sum of cubes. This is a special pattern that makes factoring certain expressions much easier. The general formula for the sum of cubes is:

a³ + b³ = (a + b)(a² - ab + b²)

This formula is your best friend when you encounter an expression that fits the sum of cubes pattern. It might look a bit intimidating at first, but let's break it down. The left side, a³ + b³, represents the sum of two terms, each raised to the power of 3 (cubed). The right side, (a + b)(a² - ab + b²), is the factored form. It consists of two factors: the first factor is the sum of the cube roots of the original terms (a + b), and the second factor is a quadratic expression (a² - ab + b²). This quadratic expression is a bit special because it's often not factorable using simple methods, but we'll get to that later. Now, why is this formula so important? Well, it provides a direct route to factoring expressions that might otherwise seem tricky. Recognizing the sum of cubes pattern allows you to quickly transform a complex expression into a product of simpler factors. This is incredibly useful in various mathematical contexts, from solving equations to simplifying algebraic expressions. So, keep this formula in your mental toolkit, guys; we're going to use it to crack our problem: 216p⁶ + 1.

Identifying Cubes in the Expression

Okay, now that we have the sum of cubes formula in our arsenal, let's see how it applies to our expression, 216p⁶ + 1. The first step is to recognize if the terms in the expression can be written as perfect cubes. Remember, a perfect cube is a number or variable that can be obtained by cubing another number or variable (raising it to the power of 3). Looking at 216p⁶, can we rewrite it as something cubed? Absolutely! 216 is 6 cubed (6 x 6 x 6 = 216), and p⁶ is (p²) cubed because when you raise a power to another power, you multiply the exponents (2 x 3 = 6). So, 216p⁶ can be rewritten as (6p²)³. What about the second term, 1? Well, 1 is a perfect cube because 1 cubed is simply 1 (1 x 1 x 1 = 1). We can write it as 1³. Now, do you see the magic happening? We've successfully expressed our original expression, 216p⁶ + 1, as the sum of two perfect cubes: (6p²)³ + 1³. This is a crucial step because it allows us to directly apply the sum of cubes formula we discussed earlier. By identifying the terms as perfect cubes, we've transformed a seemingly complex expression into a form that we can easily factor using a known pattern. In the next section, we'll plug these terms into the sum of cubes formula and see how the factoring unfolds.

Applying the Sum of Cubes Formula

Alright, we've identified that 216p⁶ + 1 can be written as (6p²)³ + 1³, which perfectly fits the sum of cubes pattern. Now, the fun part: let's apply the formula! Remember the sum of cubes formula:

a³ + b³ = (a + b)(a² - ab + b²)

In our case, a is 6p² and b is 1. We're simply going to substitute these values into the formula. First, let's find the (a + b) part. This is straightforward: (6p² + 1). Easy peasy, right? Now, let's tackle the second factor, (a² - ab + b²). This one requires a little more attention to detail, but don't worry, we'll break it down.

  • a² means squaring 6p², which gives us (6p²)² = 36p⁴.
  • -ab means multiplying 6p² and 1 and then negating the result: -(6p²)(1) = -6p².
  • b² means squaring 1, which is simply 1² = 1.

Putting it all together, the second factor (a² - ab + b²) becomes 36p⁴ - 6p² + 1. Now, we have both factors! Substituting a = 6p² and b = 1 into the sum of cubes formula, we get:

(6p²)³ + 1³ = (6p² + 1)(36p⁴ - 6p² + 1)

And there you have it! We've successfully factored 216p⁶ + 1 using the sum of cubes formula. The expression is now written as a product of two factors: (6p² + 1) and (36p⁴ - 6p² + 1). This is a significant step forward, but we're not done just yet. The next question is: can we factor these factors any further? Let's investigate in the next section.

Checking for Further Factoring

Okay, we've factored 216p⁶ + 1 into (6p² + 1)(36p⁴ - 6p² + 1). Now comes the crucial step of checking if we can factor these factors even more. Sometimes, one or both of the factors can be broken down further, leading to a complete factorization. Let's start with the first factor, (6p² + 1). Can this be factored? Well, it's a binomial (an expression with two terms), and it doesn't fit any of the common factoring patterns like the difference of squares or difference of cubes. Also, there's no common factor between the terms 6p² and 1. So, it looks like (6p² + 1) is as factored as it gets. Now, let's turn our attention to the second factor, (36p⁴ - 6p² + 1). This is a trinomial (an expression with three terms), and it might look like a quadratic expression in disguise. Remember, a quadratic expression has the general form ax² + bx + c. Our trinomial has powers of p⁴ and p², which are like x² and x if we think of p² as our variable. However, factoring this trinomial directly can be tricky. We might try to find two binomials that multiply to give this trinomial, but after some attempts, you'll likely find that it doesn't factor nicely using simple integer coefficients. In fact, this trinomial is a special type called a quadratic in form, and it often doesn't factor further using elementary techniques. So, after careful consideration, it seems that (36p⁴ - 6p² + 1) is also not factorable using standard methods. This means we've reached the end of the road in terms of factoring this expression. The factored form we obtained using the sum of cubes formula is the complete factorization.

The Final Factored Form

After our journey through the world of factoring, we've arrived at our destination! We started with the expression 216p⁶ + 1 and, using the sum of cubes formula and careful analysis, we've successfully factored it completely. So, what's the final factored form? Drumroll, please...

(6p² + 1)(36p⁴ - 6p² + 1)

This is it, guys! We've broken down the original expression into its simplest factors. The first factor, (6p² + 1), is a binomial that cannot be factored further using basic techniques. The second factor, (36p⁴ - 6p² + 1), is a trinomial that, while resembling a quadratic in form, doesn't factor nicely with integer coefficients. Therefore, this is the most factored form we can achieve. Factoring expressions like this might seem challenging at first, but by understanding key concepts like the sum of cubes and practicing your factoring skills, you'll become a pro in no time. Remember, factoring is a fundamental skill in algebra and is used extensively in higher-level mathematics. So, mastering it now will set you up for success in the future. Great job, guys, for sticking with it until the end! I hope this explanation has been clear and helpful. Keep practicing, and you'll become a factoring master!

Therefore, the correct answer is:

(6p² + 1)(36p⁴ - 6p² + 1)