Simplifying Expressions Equivalent To (3m^-2 N)^-3 / 6mn^-2

by Pedro Alvarez 60 views

Hey guys! Today, we're diving deep into the world of algebraic expressions, specifically those tricky ones involving negative exponents. We're going to break down a problem step-by-step, making sure everyone understands how to handle these types of questions. Our main goal? To master the art of simplifying expressions. So, buckle up, and let's get started!

Understanding the Problem

Before we jump into the solution, let's first understand the question. We're given the expression (3mβˆ’2n)βˆ’36mnβˆ’2\frac{\left(3 m^{-2} n\right)^{-3}}{6 m n^{-2}} and asked to find an equivalent expression from the choices provided. The key here is to remember the rules of exponents, especially how to deal with negative exponents and powers of products. We also need to keep in mind the restrictions given: mβ‰ 0m \neq 0 and nβ‰ 0n \neq 0. These are important because they prevent us from dividing by zero, which is a big no-no in math. So, let's keep these in mind as we solve this problem.

Breaking Down the Expression

To start, let's focus on the numerator of our expression: (3mβˆ’2n)βˆ’3\left(3 m^{-2} n\right)^{-3}. This part looks intimidating, but don't worry, we'll handle it! The first thing we need to remember is the power of a product rule: (ab)n=anbn(ab)^n = a^n b^n. This rule tells us that when we have a product raised to a power, we can distribute the power to each factor in the product. In our case, we have three factors inside the parentheses: 3, mβˆ’2m^{-2}, and nn. So, we need to raise each of these to the power of -3. Applying the power of a product rule helps us simplify the expression by dealing with each component separately. This is a crucial step in making the problem more manageable. Let's see how it works!

Applying the Power of a Product Rule

So, let's apply the power of a product rule to (3mβˆ’2n)βˆ’3\left(3 m^{-2} n\right)^{-3}. We get:

3βˆ’3β‹…(mβˆ’2)βˆ’3β‹…nβˆ’33^{-3} \cdot (m^{-2})^{-3} \cdot n^{-3}

Now, we have three terms to simplify. Let's start with 3βˆ’33^{-3}. Remember, a negative exponent means we take the reciprocal of the base raised to the positive exponent. So, 3βˆ’33^{-3} is the same as 133\frac{1}{3^3}, which equals 127\frac{1}{27}.

Next, we have (mβˆ’2)βˆ’3(m^{-2})^{-3}. Here, we use the power of a power rule: (am)n=amn(a^m)^n = a^{mn}. This rule tells us that when we raise a power to another power, we multiply the exponents. So, (mβˆ’2)βˆ’3(m^{-2})^{-3} becomes m(βˆ’2)(βˆ’3)m^{(-2)(-3)}, which simplifies to m6m^6. See? It's not so scary when we break it down step by step.

Finally, we have nβˆ’3n^{-3}. Again, using the rule for negative exponents, this is the same as 1n3\frac{1}{n^3}. Now we have all the pieces we need to simplify the numerator. By applying these exponent rules one at a time, we transform a complex-looking term into something much more manageable. This step-by-step approach is key to mastering these types of problems.

Simplifying the Numerator

Now, let's put the simplified terms back together. We have:

127β‹…m6β‹…1n3\frac{1}{27} \cdot m^6 \cdot \frac{1}{n^3}

Multiplying these together, we get:

m627n3\frac{m^6}{27n^3}

Great! We've simplified the numerator of our original expression. Now, let's move on to the denominator and see how it fits into the bigger picture. By focusing on one part of the expression at a time, we can avoid getting overwhelmed and make the whole process much clearer. Remember, simplifying expressions is all about taking it step by step.

Dealing with the Denominator

Okay, we've tamed the numerator, now let's turn our attention to the denominator: 6mnβˆ’26 m n^{-2}. This part looks a bit simpler, but we still need to handle that negative exponent on the nn. Remember, nβˆ’2n^{-2} is the same as 1n2\frac{1}{n^2}. So, we can rewrite the denominator as:

6mβ‹…1n2=6mn26m \cdot \frac{1}{n^2} = \frac{6m}{n^2}

Now we have a simplified denominator. Let's take a moment to appreciate how far we've come! We started with a complex expression, and by breaking it down and applying the rules of exponents, we've simplified both the numerator and the denominator. This is the essence of problem-solving in algebra – taking something complex and making it manageable. So, let's keep going and bring it all together!

Combining the Simplified Parts

Alright, we've simplified both the numerator and the denominator. Our expression now looks like this:

m627n36mn2\frac{\frac{m^6}{27n^3}}{\frac{6m}{n^2}}

This looks like a fraction within a fraction, which can be a bit daunting. But don't worry, we know how to handle this! Remember that dividing by a fraction is the same as multiplying by its reciprocal. So, we can rewrite the expression as:

m627n3Γ·6mn2=m627n3β‹…n26m\frac{m^6}{27n^3} \div \frac{6m}{n^2} = \frac{m^6}{27n^3} \cdot \frac{n^2}{6m}

Now we have a multiplication of two fractions. This is much easier to deal with! We just multiply the numerators and the denominators separately. This step is crucial because it transforms a complex division into a straightforward multiplication, making the simplification process much smoother.

Multiplying the Fractions

Let's multiply the numerators and the denominators:

m6β‹…n227n3β‹…6m\frac{m^6 \cdot n^2}{27n^3 \cdot 6m}

This gives us:

m6n2162mn3\frac{m^6 n^2}{162 m n^3}

We're getting closer! Now we have a single fraction, but we can still simplify it further. We have powers of mm and nn in both the numerator and the denominator, so we can use the quotient of powers rule to simplify. This is where our hard work really pays off, as we start to see the final simplified form emerge.

Final Simplification

Time for the final touches! We have m6n2162mn3\frac{m^6 n^2}{162 m n^3}. Let's simplify the mm terms first. We have m6m^6 in the numerator and mm (which is m1m^1) in the denominator. Using the quotient of powers rule, aman=amβˆ’n\frac{a^m}{a^n} = a^{m-n}, we get:

m6m1=m6βˆ’1=m5\frac{m^6}{m^1} = m^{6-1} = m^5

Now let's simplify the nn terms. We have n2n^2 in the numerator and n3n^3 in the denominator. Applying the same rule, we get:

n2n3=n2βˆ’3=nβˆ’1\frac{n^2}{n^3} = n^{2-3} = n^{-1}

Remember, nβˆ’1n^{-1} is the same as 1n\frac{1}{n}. So, our expression now looks like:

m5162β‹…1n\frac{m^5}{162} \cdot \frac{1}{n}

Combining these, we get:

m5162n\frac{m^5}{162n}

And there we have it! We've simplified the expression completely. It's been quite a journey, but we've made it through step by step. This final simplification brings together all the rules and techniques we've discussed, showcasing the power of methodical problem-solving.

The Answer

So, the expression (3mβˆ’2n)βˆ’36mnβˆ’2\frac{\left(3 m^{-2} n\right)^{-3}}{6 m n^{-2}} simplifies to m5162n\frac{m^5}{162 n}, which corresponds to option A. We did it! By carefully applying the rules of exponents and breaking the problem down into smaller, manageable steps, we were able to find the correct answer. Remember, practice makes perfect, so keep working on these types of problems, and you'll become a pro in no time!

Key Takeaways

Before we wrap up, let's recap the key concepts we used to solve this problem. Understanding these rules is crucial for simplifying any expression with exponents. Here’s a quick rundown:

  1. Power of a Product Rule: (ab)n=anbn(ab)^n = a^n b^n. This allows us to distribute an exponent over a product.
  2. Power of a Power Rule: (am)n=amn(a^m)^n = a^{mn}. This rule helps us simplify when an exponent is raised to another exponent.
  3. Negative Exponent Rule: aβˆ’n=1ana^{-n} = \frac{1}{a^n}. This rule shows us how to deal with negative exponents by taking the reciprocal.
  4. Quotient of Powers Rule: aman=amβˆ’n\frac{a^m}{a^n} = a^{m-n}. This rule simplifies expressions where we are dividing powers with the same base.

By mastering these rules and practicing consistently, you'll be well-equipped to tackle any exponent problem that comes your way. Keep up the great work, and remember, every complex problem can be solved by breaking it down into smaller, manageable steps!

Practice Problems

Want to test your skills further? Here are a few practice problems you can try. Remember to apply the same step-by-step approach we used in this article. Happy simplifying!

  1. Simplify: (2xβˆ’3y2)24x2yβˆ’1\frac{(2x^{-3}y^2)^2}{4x^2y^{-1}}
  2. Simplify: (5a2bβˆ’1)βˆ’310aβˆ’1b2\frac{(5a^2b^{-1})^{-3}}{10a^{-1}b^2}
  3. Simplify: (4mβˆ’1n3)βˆ’28m2nβˆ’2\frac{(4m^{-1}n^3)^{-2}}{8m^2n^{-2}}

These problems are designed to help you reinforce what you've learned and build your confidence in simplifying expressions with exponents. Give them a try, and don't hesitate to review the steps we covered if you get stuck. You've got this!

Conclusion

Simplifying expressions with exponents might seem challenging at first, but with a solid understanding of the rules and a systematic approach, you can conquer any problem. Remember, the key is to break the expression down into smaller parts, apply the rules of exponents step by step, and stay organized. With practice, you'll become more confident and efficient in simplifying these types of expressions. So keep practicing, keep learning, and most importantly, keep having fun with math! You've come a long way, and you're well on your way to mastering algebraic expressions. Keep up the awesome work!