Simplifying Algebraic Expressions: A Step-by-Step Guide

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Hey guys! Today, we're diving deep into the world of algebraic expressions, specifically focusing on how to simplify them. We'll tackle two awesome problems that will help you master this crucial math skill. So, grab your pencils and let's get started!

Problem 1: Simplifying (2xβˆ’3yβˆ’2zβˆ’4x5yβˆ’3zβˆ’3)βˆ’3\left(\frac{2x^{-3}y^{-2}z^{-4}}{x^5y^{-3}z^{-3}}\right)^{-3}

This problem might look intimidating at first, but don't worry! We're going to break it down step by step, making it super easy to understand. Our main goal here is to get rid of those negative exponents and simplify the entire expression.

Step 1: Dealing with the Innermost Fraction

First, let's focus on the fraction inside the parentheses: 2xβˆ’3yβˆ’2zβˆ’4x5yβˆ’3zβˆ’3\frac{2x^{-3}y^{-2}z^{-4}}{x^5y^{-3}z^{-3}}. The key here is to remember the rules of exponents. When dividing terms with the same base, you subtract the exponents. So, let's apply this rule to each variable:

  • For x: We have xβˆ’3x^{-3} in the numerator and x5x^5 in the denominator. Subtracting the exponents, we get xβˆ’3βˆ’5=xβˆ’8x^{-3-5} = x^{-8}.
  • For y: We have yβˆ’2y^{-2} in the numerator and yβˆ’3y^{-3} in the denominator. Subtracting the exponents, we get yβˆ’2βˆ’(βˆ’3)=yβˆ’2+3=y1=yy^{-2-(-3)} = y^{-2+3} = y^1 = y.
  • For z: We have zβˆ’4z^{-4} in the numerator and zβˆ’3z^{-3} in the denominator. Subtracting the exponents, we get zβˆ’4βˆ’(βˆ’3)=zβˆ’4+3=zβˆ’1z^{-4-(-3)} = z^{-4+3} = z^{-1}.

Now, let's rewrite the fraction with these simplified exponents:

2xβˆ’8yzβˆ’11\frac{2x^{-8}yz^{-1}}{1} or simply 2xβˆ’8yzβˆ’12x^{-8}yz^{-1}.

Step 2: Applying the Outer Exponent

Now that we've simplified the inside, let's deal with the outer exponent of -3. We have (2xβˆ’8yzβˆ’1)βˆ’3(2x^{-8}yz^{-1})^{-3}. Remember, when raising a product to a power, you raise each factor to that power. So:

  • For 2: We have 2βˆ’32^{-3}. This means 123=18\frac{1}{2^3} = \frac{1}{8}.
  • For x: We have (xβˆ’8)βˆ’3(x^{-8})^{-3}. When raising a power to a power, you multiply the exponents. So, xβˆ’8Γ—βˆ’3=x24x^{-8 \times -3} = x^{24}.
  • For y: We have yβˆ’3y^{-3}. This means 1y3\frac{1}{y^3}.
  • For z: We have (zβˆ’1)βˆ’3(z^{-1})^{-3}. Multiplying the exponents, we get zβˆ’1Γ—βˆ’3=z3z^{-1 \times -3} = z^3.

Step 3: Putting It All Together

Now, let's combine all the simplified terms:

18β‹…x24β‹…1y3β‹…z3\frac{1}{8} \cdot x^{24} \cdot \frac{1}{y^3} \cdot z^3

This simplifies to:

x24z38y3\frac{x^{24}z^3}{8y^3}

Therefore, the simplified form of (2xβˆ’3yβˆ’2zβˆ’4x5yβˆ’3zβˆ’3)βˆ’3\left(\frac{2x^{-3}y^{-2}z^{-4}}{x^5y^{-3}z^{-3}}\right)^{-3} is x24z38y3\frac{x^{24}z^3}{8y^3}.

Problem 2: Simplifying (4aβˆ’1bβˆ’2cβˆ’23abβˆ’3cβˆ’1)2\left(\frac{4a^{-1}b^{-2}c^{-2}}{3ab^{-3}c^{-1}}\right)^2

Alright, let's jump into our second problem! This one follows a similar pattern to the first, so we'll use the same exponent rules and step-by-step approach to simplify it. The key is to stay organized and take it one step at a time.

Step 1: Simplifying the Inner Fraction

We'll start by simplifying the fraction inside the parentheses: 4aβˆ’1bβˆ’2cβˆ’23abβˆ’3cβˆ’1\frac{4a^{-1}b^{-2}c^{-2}}{3ab^{-3}c^{-1}}. Just like before, we'll use the rule of subtracting exponents when dividing terms with the same base:

  • For a: We have aβˆ’1a^{-1} in the numerator and a1a^1 (or simply aa) in the denominator. Subtracting the exponents, we get aβˆ’1βˆ’1=aβˆ’2a^{-1-1} = a^{-2}.
  • For b: We have bβˆ’2b^{-2} in the numerator and bβˆ’3b^{-3} in the denominator. Subtracting the exponents, we get bβˆ’2βˆ’(βˆ’3)=bβˆ’2+3=b1=bb^{-2-(-3)} = b^{-2+3} = b^1 = b.
  • For c: We have cβˆ’2c^{-2} in the numerator and cβˆ’1c^{-1} in the denominator. Subtracting the exponents, we get cβˆ’2βˆ’(βˆ’1)=cβˆ’2+1=cβˆ’1c^{-2-(-1)} = c^{-2+1} = c^{-1}.

Don't forget the constants! We have 4 in the numerator and 3 in the denominator, so we keep the fraction 43\frac{4}{3}.

Putting it all together, our simplified fraction inside the parentheses looks like this:

4aβˆ’2bcβˆ’13\frac{4a^{-2}bc^{-1}}{3}

Step 2: Applying the Outer Exponent

Now, we need to apply the exponent of 2 to the entire expression: (4aβˆ’2bcβˆ’13)2\left(\frac{4a^{-2}bc^{-1}}{3}\right)^2. This means we'll square every factor inside the parentheses:

  • For 4/3: We have (43)2=4232=169\left(\frac{4}{3}\right)^2 = \frac{4^2}{3^2} = \frac{16}{9}.
  • For a: We have (aβˆ’2)2(a^{-2})^2. Multiplying the exponents, we get aβˆ’2Γ—2=aβˆ’4a^{-2 \times 2} = a^{-4}.
  • For b: We have b2b^2.
  • For c: We have (cβˆ’1)2(c^{-1})^2. Multiplying the exponents, we get cβˆ’1Γ—2=cβˆ’2c^{-1 \times 2} = c^{-2}.

Step 3: Cleaning Up the Expression

Let's put everything together and clean up the expression. We have:

169β‹…aβˆ’4β‹…b2β‹…cβˆ’2\frac{16}{9} \cdot a^{-4} \cdot b^2 \cdot c^{-2}

To get rid of the negative exponents, we move the terms with negative exponents to the denominator:

16b29a4c2\frac{16b^2}{9a^4c^2}

So, the simplified form of (4aβˆ’1bβˆ’2cβˆ’23abβˆ’3cβˆ’1)2\left(\frac{4a^{-1}b^{-2}c^{-2}}{3ab^{-3}c^{-1}}\right)^2 is 16b29a4c2\frac{16b^2}{9a^4c^2}.

Key Takeaways for Simplifying Expressions

Simplifying algebraic expressions can seem tricky, but with a few key strategies, you'll be a pro in no time. Here are some essential takeaways to keep in mind:

  • Master the Exponent Rules: Understanding the rules of exponents is crucial. Remember how to multiply, divide, and raise powers to powers. Knowing these rules inside and out will make simplifying expressions much smoother.
  • Deal with Negative Exponents: Negative exponents indicate reciprocals. To get rid of them, move the term to the opposite side of the fraction (numerator to denominator or vice versa) and make the exponent positive. This simple trick can significantly clean up your expressions.
  • Simplify Inside Out: When dealing with nested expressions (like the ones we tackled today), start by simplifying the innermost parts first. Work your way outwards step-by-step. This approach helps break down complex problems into smaller, more manageable chunks.
  • Stay Organized: Keep your work neat and organized. Write each step clearly and make sure you're keeping track of your exponents and coefficients. A little organization goes a long way in preventing errors.
  • Practice, Practice, Practice: Like any math skill, simplifying expressions becomes easier with practice. The more problems you solve, the more comfortable you'll become with the process and the faster you'll be able to simplify expressions.

Conclusion

There you have it! We've successfully simplified two complex algebraic expressions. Remember, the key is to take it one step at a time, apply the exponent rules correctly, and stay organized. Keep practicing, and you'll become a master of simplifying algebraic expressions in no time. Happy math-ing, guys!