Question: E = $\left(\frac{x^{-2}y^{3}z^{0}}{x^{4}y^{-1}z^{2}}\right)^{-2}$
1. **State the problem:** Simplify the expression $$E = \left(\frac{x^{-2}y^{3}z^{0}}{x^{4}y^{-1}z^{2}}\right)^{-2}$$.
2. **Recall the rules:**
- When dividing powers with the same base, subtract exponents: $$\frac{a^m}{a^n} = a^{m-n}$$.
- Any number to the zero power is 1: $$z^0 = 1$$.
- When raising a power to another power, multiply exponents: $$(a^m)^n = a^{mn}$$.
3. **Simplify inside the parentheses first:**
$$\frac{x^{-2}y^{3}z^{0}}{x^{4}y^{-1}z^{2}} = x^{-2-4} y^{3-(-1)} z^{0-2} = x^{-6} y^{4} z^{-2}$$
4. **Rewrite the expression:**
$$E = \left(x^{-6} y^{4} z^{-2}\right)^{-2}$$
5. **Apply the power of a power rule:**
$$E = x^{-6 \times (-2)} y^{4 \times (-2)} z^{-2 \times (-2)} = x^{12} y^{-8} z^{4}$$
6. **Final simplified expression:**
$$E = x^{12} \frac{z^{4}}{y^{8}}$$
This is the simplified form of the original expression.