It is given that \( y \) is directly proportional to \( (x-1)^{3} \) and \( y=20 \) when \( x=3 \). (i) Write down a formula connecting \( y \) and \( x \). (ii) Find the value of \( x \) when \( y=0.406 \), giving your answer to one decimal place.

To solve the problem, we start with the information given. (i) Since \( y \) is directly proportional to \( (x-1)^{3} \), we can express this relationship mathematically as: \[ y = k(x-1)^{3} \] where \( k \) is the constant of proportionality. We need to find the value of \( k \) using the condition that \( y = 20 \) when \( x = 3 \). Substituting these values into the equation: \[ 20 = k(3-1)^{3} \] Calculating \( (3-1)^{3} \): \[ (3-1)^{3} = 2^{3} = 8 \] Now substituting this back into the equation: \[ 20 = k \cdot 8 \] To find \( k \), we divide both sides by 8: \[ k = \frac{20}{8} = 2.5 \] Now we can write the formula connecting \( y \) and \( x \): \[ y = 2.5(x-1)^{3} \] (ii) Next, we need to find the value of \( x \) when \( y = 0.406 \). We substitute \( y \) into the formula: \[ 0.406 = 2.5(x-1)^{3} \] To isolate \( (x-1)^{3} \), we divide both sides by 2.5: \[ (x-1)^{3} = \frac{0.406}{2.5} \] Calculating the right-hand side: \[ \frac{0.406}{2.5} = 0.1624 \] Now we take the cube root of both sides to solve for \( x-1 \): \[ x-1 = \sqrt[3]{0.1624} \] Calculating the cube root: \[ x-1 \approx 0.544 \] Now, we add 1 to both sides to find \( x \): \[ x \approx 0.544 + 1 = 1.544 \] Rounding to one decimal place, we get: \[ x \approx 1.5 \] Thus, the final answers are: (i) \( y = 2.5(x-1)^{3} \) (ii) \( x \approx 1.5 \)