1. **State the problem:** Find the value of each expression: (a) $\sqrt[3]{1000} - 2^3 + 3 \times \sqrt[3]{512}$ (b) $(b - 3)^3 + \sqrt{400} - 11^2$ (assuming $b=6$ as $\sqrt[3]{343}=7$ and $\sqrt{18}$ is approximately 4.24, but since $b$ is not given, we will solve for $b=6$ to match the handwritten calculation) 2. **Recall formulas and rules:** - Cube root: $\sqrt[3]{x}$ is the number which when cubed gives $x$. - Square root: $\sqrt{x}$ is the number which when squared gives $x$. - Powers: $a^b$ means $a$ multiplied by itself $b$ times. - Order of operations: calculate powers and roots first, then multiplication and addition/subtraction. 3. **Calculate each term in (a):** - $\sqrt[3]{1000} = 10$ because $10^3 = 1000$ - $2^3 = 8$ - $\sqrt[3]{512} = 8$ because $8^3 = 512$ 4. **Substitute and simplify (a):** $$\sqrt[3]{1000} - 2^3 + 3 \times \sqrt[3]{512} = 10 - 8 + 3 \times 8$$ $$= 10 - 8 + 24$$ $$= (10 - 8) + 24 = 2 + 24 = 26$$ 5. **Calculate each term in (b) assuming $b=6$ (to match handwritten):** - $(6 - 3)^3 = 3^3 = 27$ - $\sqrt{400} = 20$ - $11^2 = 121$ 6. **Substitute and simplify (b):** $$27 + 20 - 121 = 47 - 121 = -74$$ 7. **Note:** The handwritten calculations for (b) seem inconsistent; the problem as stated cannot be solved without $b$. Using $b=6$ gives a consistent solution. **Final answers:** (a) $26$ (b) $-74$