For (ii), it’s just like a normal cubic equation, except that it’s in terms of ‘k’ instead of ‘x’.

First, we let f(k) = k³-2k²-2k+1.

To find the roots of f(k), we could use the function in our calculator to help us get the 1st root, which is -1.

This means that the 1st factor of f(k) is (k+1). For instance,
k = -1
k+1 = 0.

Presentation wise, we have to write:
By trial and error,
f(-1) = 0.

To get the quadratic factor, we divide f(k) by (k+1). It could also be done using synthetic division since the factor we are dividing is linear.

If we do it right, the quadratic factor is:
(k²-3k+1)

Hence,
(k+1)(k²-3k+1) = 0

The quadratic factor cannot be factorised using frame method/cross diagram as they are irrational. Therefore, we’ll use the quadratic formula where:
a=1, b=-3, c=1

If we substituted the respective values and simplified the expression, we should get:
(-3+√5)/2 and (-3-√5)/2