Question
Simplify, giving Solution with positive index
$\left(\frac{1}{4 a b^2 c}\right)^2 \div\left(\frac{3}{2 a^2 b^2}\right)^4$
$\left(\frac{1}{4 a b^2 c}\right)^2 \div\left(\frac{3}{2 a^2 b^2}\right)^4$
✓
Answer
$\begin{aligned} & \left(\frac{1}{4 a b^2 c}\right)^2 \div\left(\frac{3}{2 a^2 b c^2}\right)^4 \\ & =\left(\frac{1}{4 a b^2} c\right)^2 \times\left(\frac{2 a^2 b c^2}{3}\right)^4 \\ & =\frac{1^2}{4^2 a^2 b^{2 \times 2} \cdot c^2} \times \frac{2^4 a^{2 \times 4} \cdot b^4 \cdot c^{2 \times 4}}{3^4} \\ & =\frac{1^2}{3^4} \times a^{8-2} b^{4-4} c^{8-2}\left(\because 2^4=4^2\right) \\ & =\frac{1}{3 \times 3 \times 3 \times 3} a^6 b^0 c^6 \\ & =\frac{1}{81} a^6 c^6\left(\because b^0=1\right)\end{aligned}$
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