Question: e) $\left(\frac{a b^3 c^2 d^{-3}}{a^3 b^{-1} d^2}\right)^{-4} \div \left(\frac{b c^{-2} d^4}{a^{-3} b^2 c^{-2}}\right)^{-3}$
1. **State the problem:** Simplify the expression
$$\left(\frac{a b^3 c^2 d^{-3}}{a^3 b^{-1} d^2}\right)^{-4} \div \left(\frac{b c^{-2} d^4}{a^{-3} b^2 c^{-2}}\right)^{-3}$$
2. **Rewrite division as multiplication by reciprocal:**
$$= \left(\frac{a b^3 c^2 d^{-3}}{a^3 b^{-1} d^2}\right)^{-4} \times \left(\frac{b c^{-2} d^4}{a^{-3} b^2 c^{-2}}\right)^3$$
3. **Simplify inside each parenthesis first:**
For the first fraction inside the parentheses:
$$\frac{a b^3 c^2 d^{-3}}{a^3 b^{-1} d^2} = a^{1-3} b^{3 - (-1)} c^{2 - 0} d^{-3 - 2} = a^{-2} b^{4} c^{2} d^{-5}$$
For the second fraction inside the parentheses:
$$\frac{b c^{-2} d^4}{a^{-3} b^2 c^{-2}} = a^{0 - (-3)} b^{1 - 2} c^{-2 - (-2)} d^{4 - 0} = a^{3} b^{-1} c^{0} d^{4}$$
4. **Apply the exponents outside the parentheses:**
$$\left(a^{-2} b^{4} c^{2} d^{-5}\right)^{-4} = a^{8} b^{-16} c^{-8} d^{20}$$
$$\left(a^{3} b^{-1} d^{4}\right)^3 = a^{9} b^{-3} d^{12}$$
5. **Multiply the two results:**
$$a^{8} b^{-16} c^{-8} d^{20} \times a^{9} b^{-3} d^{12} = a^{8+9} b^{-16 + (-3)} c^{-8 + 0} d^{20 + 12} = a^{17} b^{-19} c^{-8} d^{32}$$
6. **Rewrite with positive exponents in denominator:**
$$= \frac{a^{17} d^{32}}{b^{19} c^{8}}$$
**Final answer:**
$$\boxed{\frac{a^{17} d^{32}}{b^{19} c^{8}}}$$