Question: Indices H.W 3
a) $\left(\frac{a^3 b^2 c^{-1}}{a^4 b^{-3} d^2}\right)^{-3} \times \left(\frac{a^{-1} b^3 d^2}{a^3 b^{-1} c^3}\right)^{-4}$
$a^{-9} b^{-6} c^3 \div a^{-12} b^9 d^{-6} \times a^{16} b^{-12} d^{-8} \div a^{-12} b^4 c^{-12}$
$\frac{a^7 b^{-18} c^3 d^{-8}}{a^{-24} b^{13} c^{-12} d^{-6}}$
$a^{31} b^{-31} c^{15} d^{-2} \to \frac{a^{31} c^{15}}{b^{31} d^{-2}}$ Ans
1. **State the problem:** Simplify the expression
$$\left(\frac{a^3 b^2 c^{-1}}{a^4 b^{-3} d^2}\right)^{-3} \times \left(\frac{a^{-1} b^3 d^2}{a^3 b^{-1} c^3}\right)^{-4}$$
2. **Recall exponent rules:**
- $\left(\frac{x}{y}\right)^n = \frac{x^n}{y^n}$
- $x^{-n} = \frac{1}{x^n}$
- When multiplying like bases, add exponents: $a^m \times a^n = a^{m+n}$
- When dividing like bases, subtract exponents: $\frac{a^m}{a^n} = a^{m-n}$
3. **Simplify inside each parenthesis:**
$$\frac{a^3 b^2 c^{-1}}{a^4 b^{-3} d^2} = a^{3-4} b^{2-(-3)} c^{-1} d^{-2} = a^{-1} b^{5} c^{-1} d^{-2}$$
$$\frac{a^{-1} b^3 d^2}{a^3 b^{-1} c^3} = a^{-1-3} b^{3-(-1)} d^{2} c^{-3} = a^{-4} b^{4} d^{2} c^{-3}$$
4. **Apply the outer exponents:**
$$\left(a^{-1} b^{5} c^{-1} d^{-2}\right)^{-3} = a^{-1 \times -3} b^{5 \times -3} c^{-1 \times -3} d^{-2 \times -3} = a^{3} b^{-15} c^{3} d^{6}$$
$$\left(a^{-4} b^{4} d^{2} c^{-3}\right)^{-4} = a^{-4 \times -4} b^{4 \times -4} d^{2 \times -4} c^{-3 \times -4} = a^{16} b^{-16} d^{-8} c^{12}$$
5. **Multiply the two results:**
$$a^{3} b^{-15} c^{3} d^{6} \times a^{16} b^{-16} d^{-8} c^{12} = a^{3+16} b^{-15-16} c^{3+12} d^{6-8} = a^{19} b^{-31} c^{15} d^{-2}$$
6. **Final simplified expression:**
$$\frac{a^{19} c^{15}}{b^{31} d^{2}}$$
**Note:** The user's final answer was $\frac{a^{31} c^{15}}{b^{31} d^{-2}}$, but the correct exponent for $a$ is $19$ and $d^{-2}$ in denominator is $d^{2}$.
Hence, the simplified expression is:
$$\boxed{\frac{a^{19} c^{15}}{b^{31} d^{2}}}$$