Simplifying The Expression: $x^3 \sqrt[4]{32 X^5 Y^3}$

Hey guys! Today, we're diving into the world of algebra to tackle a simplification problem. Our mission, should we choose to accept it (and we do!), is to simplify the expression $x^3 \sqrt[4]{32 x^5 y^3}$. This might look a bit intimidating at first glance, but don't worry, we'll break it down step by step and make it super easy to understand. We'll explore the fundamental principles of simplifying radicals and exponents. By the end of this guide, you'll not only know how to solve this specific problem but also have a solid foundation for tackling similar algebraic challenges. Ready to jump in? Let's get started!

Breaking Down the Expression

Alright, let's get our hands dirty with this expression. The key to simplifying expressions like this is to identify the components we can work with individually. We have $x^3$ hanging out on the outside, and then we have a fourth root, $\sqrt[4]{}$, containing $32x^5y^3$. To make things easier, let's focus on what's inside the radical first. Think of it like decluttering your room – you tackle one section at a time!

When simplifying radicals, the main goal is to find perfect powers that we can "pull out" of the radical. In this case, we're dealing with a fourth root, so we're looking for factors that can be written as something to the fourth power. This is where understanding exponents and their properties comes in handy. Remember, $\sqrt[n]{a^n} = a$, assuming $a$ is non-negative. So, let's see how we can apply this to our expression. We'll start by breaking down the number 32 and the variables $x^5$ and $y^3$ to see if we can find any perfect fourth powers hiding within them. It's like being a detective, searching for clues to unlock the simplified form.

Prime Factorization of 32

Let's start with the numerical part, 32. To find perfect fourth powers, we need to break 32 down into its prime factors. Prime factorization is like dismantling a machine into its smallest components to understand how it works. We can express 32 as a product of prime numbers. So, 32 can be written as $2 imes 2 imes 2 imes 2 imes 2$, which is $2^5$. Now we can rewrite $2^5$ as $2^4 imes 2$. Aha! We've found a perfect fourth power: $2^4$. This is fantastic news because we can take the fourth root of $2^4$, which will simply be 2. See how breaking it down made it easier? This is the magic of prime factorization.

Simplifying Variable Exponents

Next up, let's tackle the variables inside the radical: $x^5$ and $y^3$. Remember, we're looking for powers of 4 since we have a fourth root. For $x^5$, we can rewrite it as $x^4 imes x$. Just like with 32, we've identified a perfect fourth power, $x^4$. When we take the fourth root of $x^4$, we'll get $x$. Things are starting to shape up nicely! Now let's consider $y^3$. Can we rewrite $y^3$ in terms of a fourth power? Unfortunately, no. The exponent 3 is smaller than 4, so we can't extract any whole powers of $y$ from the fourth root. This means $y^3$ will have to stay inside the radical for now. Don't worry, though; it's still a part of the puzzle, and we'll deal with it accordingly.

Rewriting the Expression

Okay, we've done some serious detective work here. We've broken down 32 into $2^4 imes 2$ and $x^5$ into $x^4 imes x$. Now, let's rewrite the entire expression inside the radical using these simplified forms. Our original radical expression, $\sqrt[4]{32 x^5 y^3}$, can now be written as $\sqrt[4]{2^4 imes 2 imes x^4 imes x imes y^3}$. This might look a bit long, but we're actually in a great position to simplify it. By regrouping the terms, we've made the perfect fourth powers stand out, ready to be extracted from the radical. We're essentially preparing the expression for its final transformation, like a caterpillar getting ready to become a butterfly. The next step will be to actually take the fourth root of the terms we've identified, and that's where the real magic happens. Are you excited? I know I am!

Extracting Perfect Fourth Roots

Alright, we've successfully rewritten our expression inside the radical as $\sqrt[4]{2^4 imes 2 imes x^4 imes x imes y^3}$. Now comes the really fun part: extracting the perfect fourth roots! Remember, the fourth root of a number raised to the fourth power is simply that number. So, $\sqrt[4]{2^4}$ is 2, and $\sqrt[4]{x^4}$ is $x$ (assuming $x$ is non-negative). It's like unlocking a secret code – we're using the properties of radicals and exponents to simplify the expression.

Let's take those perfect fourth roots out of the radical. We have $2^4$ and $x^4$, so we can pull out a 2 and an $x$. When we do this, they'll multiply with the $x^3$ that's already sitting outside the radical. This is where things start to come together nicely. The terms that don't have a perfect fourth power – namely 2, $x$, and $y^3$ – will remain inside the radical. They're like the core elements that couldn't be simplified further, but they're still essential to the expression.

Combining Terms Outside the Radical

Okay, so we've pulled out a 2 and an $x$ from the fourth root. Now we need to combine these with the $x^3$ that's already outside the radical. We have $x^3$ multiplied by $x$, which gives us $x^4$. Remember the rule for multiplying exponents with the same base: you add the exponents. So, $x^3 imes x^1 = x^{3+1} = x^4$. This is a crucial step in simplifying the expression. We're essentially consolidating the terms outside the radical to make the expression cleaner and more manageable. It's like organizing your tools after a big project – you put everything in its place so you can easily find it next time.

The Remaining Terms Inside the Radical

Now, let's talk about what's left inside the radical. We had $\sqrt[4]{2^4 imes 2 imes x^4 imes x imes y^3}$, and we've taken out the $2^4$ and $x^4$. This leaves us with $\sqrt[4]{2 imes x imes y^3}$, or simply $\sqrt[4]{2xy^3}$. These are the terms that couldn't be simplified further in terms of fourth powers. They're the irreducible components of the radical expression. It's like the last few pieces of a puzzle that fit perfectly together but can't be broken down any further. So, we'll keep them as they are inside the radical.

By carefully extracting the perfect fourth roots and combining the terms outside the radical, we've made significant progress in simplifying our original expression. We've essentially separated the parts that could be simplified from those that couldn't, making the expression much easier to handle. The next step is to put it all together and write out the fully simplified expression. Are you ready to see the final result? Let's do it!

The Simplified Expression

We've done it! We've successfully broken down the original expression, extracted the perfect fourth roots, and combined the terms outside the radical. Now, let's put all the pieces together to reveal the fully simplified expression.

We had $x^3 \sqrt[4]{32 x^5 y^3}$. We simplified the radical part to $\sqrt[4]{2^4 imes 2 imes x^4 imes x imes y^3}$. We extracted the fourth roots of $2^4$ and $x^4$, which gave us 2 and $x$, respectively. We then multiplied the $x$ by the $x^3$ outside the radical, resulting in $x^4$. The terms left inside the radical were 2, $x$, and $y^3$, giving us $\sqrt[4]{2xy^3}$.

So, putting it all together, our simplified expression is $2x^4 \sqrt[4]{2xy^3}$. Ta-da! We've taken a complex-looking expression and transformed it into a much simpler, more elegant form. This is the power of understanding the properties of radicals and exponents. It's like having a secret decoder ring that allows you to unlock the hidden simplicity within mathematical expressions.

Comparing with the Options

Now, let's compare our simplified expression, $2x^4 \sqrt[4]{2xy^3}$, with the options provided:

A. $4 x^3 \sqrt[4]{2 x^2 y^3}$ B. $2 x^4 \sqrt[4]{2 x y^3}$ C. $2 x^4 y \sqrt[4]{4 x y^3}$ D. $2 x^4 \sqrt[4]{4 x y^3}$

Looking at the options, we can see that option B, $2 x^4 \sqrt[4]{2 x y^3}$, matches our simplified expression perfectly! This confirms that we've successfully simplified the expression and found the correct answer. It's always satisfying when your hard work pays off, isn't it? By methodically breaking down the problem and applying the rules of algebra, we were able to navigate through the complexity and arrive at the correct solution.

Conclusion

Great job, guys! We've successfully simplified the expression $x^3 \sqrt[4]{32 x^5 y^3}$ to $2 x^4 \sqrt[4]{2 x y^3}$. We walked through the entire process, from breaking down the expression inside the radical to extracting perfect fourth roots and combining terms. Along the way, we reinforced key concepts such as prime factorization, exponent rules, and the properties of radicals. This problem is a fantastic example of how understanding the fundamentals of algebra can empower you to tackle even seemingly complex challenges.

Remember, the key to simplifying expressions like this is to take it one step at a time. Don't be intimidated by the initial appearance. Break the problem down into smaller, more manageable parts. Identify the perfect powers, extract them from the radical, and combine the terms. With practice and a solid understanding of the underlying principles, you'll become a pro at simplifying radical expressions. Keep up the great work, and I'll see you in the next algebra adventure!

So, to recap, the correct answer is:

B. $2 x^4 \sqrt[4]{2 x y^3}$