1. **State the problem:** Find the first 4 terms of the expansion of $ (1-8x)^3 $ in ascending powers of $ x $, then use $ x=\frac{1}{100} $ to evaluate $ \sqrt[3]{23} $ correct to 5 significant figures. 2. **Formula used:** Use the binomial expansion for $ (1+u)^n = \sum_{k=0}^n \binom{n}{k} u^k $ where $ n=3 $ and $ u = -8x $. 3. **Expand:** $$ (1-8x)^3 = \binom{3}{0} (1)^3 (-8x)^0 + \binom{3}{1} (1)^2 (-8x)^1 + \binom{3}{2} (1)^1 (-8x)^2 + \binom{3}{3} (1)^0 (-8x)^3 $$ 4. **Calculate each term:** - $ \binom{3}{0} = 1 $, term: $ 1 $ - $ \binom{3}{1} = 3 $, term: $ 3 \times (-8x) = -24x $ - $ \binom{3}{2} = 3 $, term: $ 3 \times (-8x)^2 = 3 \times 64x^2 = 192x^2 $ - $ \binom{3}{3} = 1 $, term: $ 1 \times (-8x)^3 = -512x^3 $ 5. **Write the expansion:** $$ (1-8x)^3 = 1 - 24x + 192x^2 - 512x^3 $$ 6. **Use $ x=\frac{1}{100} $ to approximate $ \sqrt[3]{23} $:** Note that $ 23 = 27 \times \left(1 - \frac{4}{27}\right) $, so $ \sqrt[3]{23} = 3 \times (1 - \frac{4}{27})^{1/3} $. 7. **Set $ x = \frac{4}{27} \approx 0.148148 $ and use the binomial expansion for $ (1 - x)^{1/3} $:** The binomial expansion for fractional powers is: $$ (1 - x)^{1/3} = 1 - \frac{1}{3}x + \frac{1}{9}x^2 - \frac{5}{81}x^3 + \cdots $$ 8. **Calculate terms:** - $ 1 $ - $ -\frac{1}{3} \times 0.148148 = -0.0493827 $ - $ \frac{1}{9} \times (0.148148)^2 = \frac{1}{9} \times 0.02193 = 0.002436 $ - $ -\frac{5}{81} \times (0.148148)^3 = -\frac{5}{81} \times 0.00325 = -0.0002007 $ 9. **Sum terms:** $$ 1 - 0.0493827 + 0.002436 - 0.0002007 = 0.9528526 $$ 10. **Multiply by 3:** $$ 3 \times 0.9528526 = 2.85856 $$ 11. **Final answer:** $$ \sqrt[3]{23} \approx 2.8586 \quad \text{(to 5 significant figures)} $$