Like Radicals Of ∛54: A Step-by-Step Solution
Hey there, math enthusiasts! Today, we're diving into the world of radicals, specifically cube roots, and tackling the question: Which radical is like $\sqrt[3]{54}$ after simplifying? To nail this, we'll simplify each radical and see which one matches our simplified version of $\sqrt[3]{54}$. Buckle up, because we're about to make radicals a piece of cake!
Understanding Like Radicals
Before we jump into simplifying, let's quickly recap what like radicals actually are. Like radicals are radicals that have the same index (the little number in the crook of the radical symbol, like the 3 in a cube root) and the same radicand (the number inside the radical symbol). For example, 2$\sqrt{3}$ and 5$\sqrt{3}$ are like radicals because they both have a square root (index of 2) and the radicand is 3. However, $\sqrt{2}$ and $\sqrt[3]{2}$ are not like radicals because they have different indices, even though the radicand is the same. Similarly, $\sqrt{5}$ and $\sqrt{7}$ are not like radicals because they have the same index (2) but different radicands. Understanding this concept is crucial for simplifying and comparing radicals.
When we talk about simplifying radicals, we're essentially trying to express them in their simplest form, where the radicand has no perfect square factors (for square roots), perfect cube factors (for cube roots), and so on. This often involves factoring the radicand and pulling out any perfect powers. Once we've simplified, it becomes much easier to identify like radicals. So, remember, the key is to simplify first and then compare the indices and radicands. This foundation will help us confidently tackle the question at hand and any other radical simplification problems that come our way. Let's keep this in mind as we move forward and simplify each of the given radicals.
Simplifying $\sqrt[3]{54}$
Let's start with the given radical, $\sqrt[3]{54}$. Our goal here is to break down 54 into its prime factors and see if we can find any perfect cubes hiding in there. Factoring 54, we get 54 = 2 × 27. Ah, 27! That's a perfect cube because 27 = 3 × 3 × 3 = 3³. So, we can rewrite our radical as $\sqrt[3]{54}$ = $\sqrt[3]{2 × 27}$ = $\sqrt[3]{2 × 3^3}$. Now, we can use the property of radicals that says $\sqrt[n]{a × b}$ = $\sqrt[n]{a}$ × $\sqrt[n]{b}$. Applying this, we get $\sqrt[3]{2 × 3^3}$ = $\sqrt[3]{3^3}$ × $\sqrt[3]{2}$. And finally, since the cube root of 3³ is simply 3, we have $\sqrt[3]{3^3}$ × $\sqrt[3]{2}$ = 3$\sqrt[3]{2}$. So, the simplified form of $\sqrt[3]{54}$ is 3$\sqrt[3]{2}$.
This simplified form, 3$\sqrt[3]{2}$, is our benchmark. We're looking for another radical that, when simplified, will also have a radicand of 2 under a cube root. Remember, the coefficient (the 3 in 3$\sqrt[3]{2}$) doesn't matter when we're identifying like radicals; it's all about the index and the radicand. Now, let's move on to simplifying the other radicals provided in the question. We'll apply the same process of prime factorization and looking for perfect cubes to each one. By comparing their simplified forms to 3$\sqrt[3]{2}$, we'll be able to pinpoint the like radical. Keep this step-by-step approach in mind, and you'll be simplifying radicals like a pro in no time! So, let's dive into the next radical and continue our simplification journey.
Simplifying $\sqrt[3]{24}$
Next up, we have $\sqrt[3]{24}$. Let's simplify this one using the same method we used for $\sqrt[3]{54}$. We need to factor 24 and see if there are any perfect cubes lurking within. Factoring 24, we get 24 = 2 × 12 = 2 × 2 × 6 = 2 × 2 × 2 × 3 = 2³ × 3. See that? We've got a 2³ in there, which is a perfect cube! So, we can rewrite $\sqrt[3]{24}$ as $\sqrt[3]{2^3 × 3}$. Now, using our radical property $\sqrt[n]{a × b}$ = $\sqrt[n]{a}$ × $\sqrt[n]{b}$, we can separate this into $\sqrt[3]{2^3}$ × $\sqrt[3]{3}$. The cube root of 2³ is simply 2, so we have 2$\sqrt[3]{3}$.
Okay, let's pause and compare this to our simplified form of $\sqrt[3]{54}$, which was 3$\sqrt[3]{2}$. Notice that $\sqrt[3]{24}$ simplifies to 2$\sqrt[3]{3}$. The radicand here is 3, while the radicand in our benchmark is 2. Even though both are cube roots (same index), the different radicands mean that 2$\sqrt[3]{3}$ and 3$\sqrt[3]{2}$ are not like radicals. So, $\sqrt[3]{24}$ is not the radical we're looking for. But don't worry, we're not giving up! We've still got more radicals to simplify. The key here is to keep practicing this process of factoring and looking for perfect powers. Each simplification we do gets us closer to finding our like radical. Let's move on to the next option and see if it matches our benchmark. Remember, we're hunting for a simplified radical with a cube root and a radicand of 2.
Simplifying $\sqrt[3]{162}$
Now let's tackle $\sqrt[3]{162}$. As always, we start by factoring 162 to see if we can uncover any perfect cubes. Factoring 162, we get 162 = 2 × 81 = 2 × 9 × 9 = 2 × 3 × 3 × 3 × 3 = 2 × 3⁴. We can rewrite 3⁴ as 3³ × 3. So, 162 = 2 × 3³ × 3. Now we can express our radical as $\sqrt[3]{162}$ = $\sqrt[3]{2 × 3^3 × 3}$. Time to use our radical property again! We have $\sqrt[3]{2 × 3^3 × 3}$ = $\sqrt[3]{3^3}$ × $\sqrt[3]{2 × 3}$. The cube root of 3³ is 3, so we have 3$\sqrt[3]{2 × 3}$ = 3$\sqrt[3]{6}$.
Let's take a moment to compare this to our simplified benchmark, 3$\sqrt[3]2}$. We've simplified $\sqrt[3]{162}$ to 3$\sqrt[3]{6}$. Both have the same index (3, meaning they're cube roots), but the radicands are different$ and 3$\sqrt[3]{2}$ are not like radicals. $\sqrt[3]{162}$ is not our match. Don't be discouraged! We're learning valuable skills in simplifying radicals, and with each one we do, we get better at spotting those perfect cubes. We have two more radicals to check, so let's keep up the momentum. Remember, we're on the lookout for a cube root with a radicand of 2. Let's move on to the next radical and continue our quest!
Simplifying $\sqrt{128}$
Alright, let's move on to $\sqrt{128}$. Notice anything different about this radical compared to the others we've worked with so far? This one is a square root, not a cube root! This is a major clue. Remember, for radicals to be
