Question: = 3 1^-1
(b) 16^(-3/2)
2 FIRST TERM TEST/ GRADE 8/ MA
1. **State the problem:** Simplify the expression $$16^{-\frac{3}{2}}$$.
2. **Recall the rules:** For any positive number $a$ and rational exponent $m/n$, $$a^{\frac{m}{n}} = \sqrt[n]{a^m} = \left(\sqrt[n]{a}\right)^m$$.
Also, a negative exponent means reciprocal: $$a^{-k} = \frac{1}{a^k}$$.
3. **Apply the negative exponent rule:**
$$16^{-\frac{3}{2}} = \frac{1}{16^{\frac{3}{2}}}$$
4. **Rewrite the positive fractional exponent:**
$$16^{\frac{3}{2}} = \left(16^{\frac{1}{2}}\right)^3$$
5. **Calculate the square root:**
$$16^{\frac{1}{2}} = \sqrt{16} = 4$$
6. **Raise to the power 3:**
$$4^3 = 64$$
7. **Combine all steps:**
$$16^{-\frac{3}{2}} = \frac{1}{64}$$
**Final answer:** $$\boxed{\frac{1}{64}}$$