Simplify the expression: $\frac{(2x^3y^1z^{-2})^{-2}x^4y^8z^{-2}}{5x^5y^4z^2}$

1. Problem Description

Simplify the expression:

$\frac{(2x^3y^1z^{-2})^{-2}x^4y^8z^{-2}}{5x^5y^4z^2}$

2. Solution Steps

First, we apply the power of a product rule to the numerator's first term:

$(ab)^n = a^nb^n$

$(2x^3y^1z^{-2})^{-2} = 2^{-2}(x^3)^{-2}(y^1)^{-2}(z^{-2})^{-2} = 2^{-2}x^{-6}y^{-2}z^4$

So the expression becomes:

$\frac{2^{-2}x^{-6}y^{-2}z^4x^4y^8z^{-2}}{5x^5y^4z^2}$

Next, we combine like terms in the numerator by using the rule $a^m a^n = a^{m+n}$:

$x^{-6}x^4 = x^{-6+4} = x^{-2}$

$y^{-2}y^8 = y^{-2+8} = y^6$

$z^4z^{-2} = z^{4+(-2)} = z^2$

The expression is now:

$\frac{2^{-2}x^{-2}y^6z^2}{5x^5y^4z^2}$

Now, we use the rule $\frac{a^m}{a^n} = a^{m-n}$ to simplify:

$\frac{x^{-2}}{x^5} = x^{-2-5} = x^{-7}$

$\frac{y^6}{y^4} = y^{6-4} = y^2$

$\frac{z^2}{z^2} = z^{2-2} = z^0 = 1$

$2^{-2} = \frac{1}{2^2} = \frac{1}{4}$

The expression is now:

$\frac{\frac{1}{4}x^{-7}y^2(1)}{5}$

Simplifying the fraction gives:

$\frac{x^{-7}y^2}{4 \cdot 5} = \frac{x^{-7}y^2}{20}$

Since $x^{-7} = \frac{1}{x^7}$, we can rewrite this as:

$\frac{y^2}{20x^7}$

3. Final Answer

$\frac{y^2}{20x^7}$