- Evaluate constants: \((-2)^3 = -8\) and \((3)^2 = 9\).
- Apply exponent properties to simplify variable terms: \(\frac{x^7}{x^2} = x^5\), \(\frac{y^4}{y^7} = y^{-3} = \frac{1}{y^3}\), and \(\frac{z^4}{z^3} = z\).
- Combine all parts into a single fraction: \(\frac{-8 x^5 z}{9 y^3}\).
- The final simplified expression is \(\boxed{\frac{-8x^5z}{9y^3}}\).
Explanation
- Analysis of the Expression
We aim to simplify the algebraic expression:
\[ \frac{(-2)^3 x^7 (y z)^4}{(3)^2 x^2 y^7 z^3} \]
Our strategy is to handle the numerical coefficients and the variables separately, then use exponent rules to simplify the expression into a form where all exponents are positive. Recall the rules:
- Power of a product: \((ab)^n = a^n b^n\)
- Quotient of powers: \(\frac{a^m}{a^n} = a^{m-n}\)
- Negative exponents: \(a^{-n} = \frac{1}{a^n}\)
- Numerical and Algebraic Expansion
First, we calculate the numerical constants:
- The numerator contains \((-2)^3 = -8\).
- The denominator contains \((3)^2 = 9\).
Next, we expand the term \((yz)^4\) using the power of a product rule, which gives us \(y^4 z^4\). Substituting these back, the expression becomes:
\[ \frac{-8 \cdot x^7 \cdot y^4 \cdot z^4}{9 \cdot x^2 \cdot y^7 \cdot z^3} \]
- Simplifying Exponents and Final Result
Now we apply the quotient rule \(\frac{a^m}{a^n} = a^{m-n}\) to each variable:
- For \(x\): \(x^{7-2} = x^5\)
- For \(y\): \(y^{4-7} = y^{-3} = \frac{1}{y^3}\)
- For \(z\): \(z^{4-3} = z^1 = z\)
Combining these, we get:
\[ \frac{-8}{9} \cdot x^5 \cdot y^{-3} \cdot z^1 = \frac{-8 x^5 z}{9 y^3} \]
Since the question asks for only positive exponents, the negative power of \(y\) is moved to the denominator, ensuring all exponents in our final expression are positive.
Examples
In fields like physics or chemistry, formulas often involve complex products and ratios of variables, such as calculating the concentration of a chemical or the force between particles. Simplifying these expressions is essential for easier interpretation. For example, when dealing with the expression for density \(\rho = \frac{m}{V}\) where \(V\) is a product of dimensions, simplifying the algebra ensures you can quickly determine how changing one dimension (like length \(L\)) affects the overall density \(\rho \propto L^{-k}\). This practice makes complex scientific models much more manageable.