Simplify: Cube Root Sum Of 125x^10y^13 + 27x^10y^13

Hey there, math enthusiasts! Today, we're diving into an intriguing problem that involves simplifying the sum of cube roots. It looks a bit intimidating at first glance, but fear not! With a dash of algebraic magic and a sprinkle of understanding cube roots, we'll break it down step by step. So, let's jump right into the heart of the problem and unravel the mystery behind this mathematical expression: √[3]{125 x^{10} y^{13}} + √[3]{27 x^{10} y^{13}}.

Decoding Cube Roots: The Foundation of Our Solution

Before we tackle the main problem, let's quickly brush up on what cube roots are all about. You see, a cube root of a number is a value that, when multiplied by itself three times, gives you the original number. Think of it like this: the cube root of 8 is 2 because 2 * 2 * 2 = 8. We write the cube root using the radical symbol with a small '3' nestled in the crook: √[3]. Understanding this concept is crucial because we'll be using it extensively to simplify the terms within our expression. The key takeaway here is to identify perfect cubes within the radicand (the expression under the radical symbol). For example, in √[3]{125}, 125 is a perfect cube (5 * 5 * 5), so we can easily find its cube root.

Now, let's extend this idea to variables with exponents. When taking the cube root of a variable raised to a power, we divide the exponent by 3. For instance, √[3]{x^6} = x^(6/3) = x^2. If the exponent isn't perfectly divisible by 3, we'll need to get a tad creative and rewrite the expression to isolate the largest possible multiple of 3. This is precisely what we'll do when dealing with terms like x^{10} and y^{13} in our problem. Remember, guys, the goal is to extract as much as we can from under the cube root to simplify the expression.

Dissecting the Terms: A Step-by-Step Approach

Okay, with our cube root knowledge in place, let's dissect the terms in our sum one by one. We have two terms: √[3]{125 x^{10} y^{13}} and √[3]{27 x^{10} y^{13}}. Let's start with the first term, √[3]{125 x^{10} y^{13}}. Our mission is to simplify this expression by identifying perfect cubes and extracting them from under the cube root. First, let's focus on the coefficient, 125. As we discussed earlier, 125 is a perfect cube (5 * 5 * 5 = 125), so its cube root is simply 5. That's a great start! Now, let's move on to the variables.

We have x^{10}. Since 10 isn't perfectly divisible by 3, we need to rewrite it as the sum of the largest multiple of 3 less than 10 and the remainder. In this case, 10 = 9 + 1. So, we can rewrite x^{10} as x^9 * x^1. Why do we do this? Because x^9 is a perfect cube (√[3]{x^9} = x^(9/3) = x^3). The remaining x^1 will stay under the cube root. Next up, we have y^{13}. Similarly, 13 isn't divisible by 3, so we rewrite it as 12 + 1. Thus, y^{13} becomes y^{12} * y^1. Again, y^{12} is a perfect cube (√[3]{y^{12}} = y^(12/3) = y^4), and the remaining y^1 stays under the cube root. Now, let's put it all together. We can rewrite √[3]{125 x^{10} y^{13}} as √[3]{125 * x^9 * x * y^{12} * y}. Taking the cube root of the perfect cubes, we get 5 * x^3 * y^4 * √[3]{x * y}. Phew! That's the first term simplified. Let's tackle the second term with the same approach.

The Second Term: A Mirror Image of Simplification

Now, let's turn our attention to the second term, √[3]{27 x^{10} y^{13}}. The process will be very similar to what we did with the first term. We'll break it down step by step, identifying perfect cubes and extracting them from the cube root. First, let's look at the coefficient, 27. 27 is also a perfect cube (3 * 3 * 3 = 27), so its cube root is 3. Awesome! We're making progress.

The variable terms, x^10} and y^{13}, are the same as in the first term. So, we already know how to handle them. We rewrite x^{10} as x^9 * x and y^{13} as y^{12} * y. As before, x^9 and y^{12} are perfect cubes. Now, let's rewrite the entire term √[3]{27 x^{10 y^{13}} = √[3]{27 * x^9 * x * y^{12} * y}. Taking the cube root of the perfect cubes, we get 3 * x^3 * y^4 * √[3]{x * y}. Great! We've simplified the second term as well.

The Grand Finale: Combining the Simplified Terms

We've reached the final stage! We've simplified both terms in our original sum, and now it's time to combine them. Remember, our original problem was √[3]{125 x^{10} y^{13}} + √[3]{27 x^{10} y^{13}}. We simplified the first term to 5 * x^3 * y^4 * √[3]{x * y} and the second term to 3 * x^3 * y^4 * √[3]{x * y}. Now we can rewrite the sum as:

5 * x^3 * y^4 * √[3]{x * y} + 3 * x^3 * y^4 * √[3]{x * y}

Notice something? Both terms have a common factor: x^3 * y^4 * √[3]{x * y}. This means we can factor it out, just like we do with regular algebraic expressions. Factoring out the common factor, we get:

(5 + 3) * x^3 * y^4 * √[3]{x * y}

Now, we simply add the numbers inside the parentheses: 5 + 3 = 8. So, our final simplified expression is:

8x^3y4√[3]{xy}

And there you have it! We've successfully simplified the sum of cube roots. It might have seemed like a daunting task at first, but by breaking it down into smaller steps, understanding cube roots, and identifying perfect cubes, we were able to conquer the problem. Remember, guys, math is all about taking things one step at a time and applying the right concepts. Keep practicing, and you'll be simplifying even the most complex expressions in no time!

Key Takeaways and Further Exploration

Let's recap the key takeaways from this problem. First, understanding the concept of cube roots and perfect cubes is essential for simplification. Second, when dealing with variables raised to powers under a cube root, we look for the largest multiple of 3 that is less than or equal to the exponent. This allows us to extract the variable with a whole number exponent from the cube root. Third, factoring out common factors is a powerful technique for simplifying expressions, especially after simplifying individual terms.

If you're eager to explore further, you can try similar problems with different coefficients and exponents. You can also investigate simplifying higher-order roots, such as fourth roots or fifth roots. The principles remain the same, but the numbers and exponents might be a bit larger. Another interesting avenue to explore is the simplification of expressions involving multiple cube roots or nested radicals. These types of problems can be a bit more challenging, but they offer a great opportunity to deepen your understanding of radicals and algebraic manipulation. Keep challenging yourself, and you'll become a true math whiz!

So, guys, I hope this journey into simplifying cube roots has been both enlightening and enjoyable. Remember, math is not just about numbers and equations; it's about problem-solving, logical thinking, and the thrill of discovery. Keep exploring, keep questioning, and keep simplifying! Until next time, happy calculating!