Between which two integers do the following surds lie? \( \begin{array}{ll}\text { a) } \sqrt[3]{6} & \text { b) } \sqrt[3]{-45} \\ \text { c) } \sqrt[3]{78} & \text { d) }-\sqrt[3]{130}\end{array} \)

Let's determine between which two integers each of the given cube roots lies. ### a) \( \sqrt[3]{6} \) We need to find integers \( n \) such that: \[ n^3 < 6 < (n+1)^3 \] - \( 1^3 = 1 \) - \( 2^3 = 8 \) Since \( 1 < 6 < 8 \), it follows that: \[ 1 < \sqrt[3]{6} < 2 \] ### b) \( \sqrt[3]{-45} \) For negative numbers, the cube root will also be negative. We find integers \( n \) such that: \[ n^3 < -45 < (n+1)^3 \] - \( (-4)^3 = -64 \) - \( (-3)^3 = -27 \) Since \( -64 < -45 < -27 \), it follows that: \[ -4 < \sqrt[3]{-45} < -3 \] ### c) \( \sqrt[3]{78} \) Find integers \( n \) such that: \[ n^3 < 78 < (n+1)^3 \] - \( 4^3 = 64 \) - \( 5^3 = 125 \) Since \( 64 < 78 < 125 \), it follows that: \[ 4 < \sqrt[3]{78} < 5 \] ### d) \( -\sqrt[3]{130} \) First, find the cube root of 130: \[ n^3 < 130 < (n+1)^3 \] - \( 5^3 = 125 \) - \( 6^3 = 216 \) Thus: \[ 5 < \sqrt[3]{130} < 6 \] Negating this: \[ -6 < -\sqrt[3]{130} < -5 \] ### Summary - **a)** \( \sqrt[3]{6} \) lies between **1 and 2** - **b)** \( \sqrt[3]{-45} \) lies between **-4 and -3** - **c)** \( \sqrt[3]{78} \) lies between **4 and 5** - **d)** \( -\sqrt[3]{130} \) lies between **-6 and -5** **Answer:** a) 1 and 2; b) –4 and –3; c) 4 and 5; d) –6 and –5.