For (i), since the 3 roots are 1, k and k²,

x = 1
x-1 = 0

x= k
x-k = 0

x = k²
x-k² = 0

If we work backwards and combine them,
(x-1)(x-k)(x-k²) = 0

There may or may not be any constant in front of the brackets. However, given the coefficient of x³ is -1, we know that there should be a “-1” there.

Hence,
f(x) = -(x-1)(x-k)(x-k²)

Then, we find the zero of “x-2”.
x-2 = 0
x = 2

Using remainder theorem, when x=2, we get a reminder of -3. Therefore, we substitute x=2 and make f(2) equal to -3.
f(2) = -(2-1)(2-k)(2-k²)
-3 = -(1)(2-k)(2-k²)

Divide/Multiply both sides by -1,
3 = (1)(2-k)(2-k²)
(2-k)(2-k²) = 3
4-2k²-2k+k³ = 3
k³-2k²-2k+4 = 3
k³-2k²-2k+4-3 = 0
k³-2k²-2k+1 = 0 (shown)