Actually, this part is usually an algebraic solving approach in replacement of a graphical solving approach in part (ii).

We must have been given the equations of the curve and the line beforehand to even do part (ii). We need these equations to do this question. Since I do not see the question at all here, I'm just going to draft out a random question.

Suppose the curve equation is y = x³ - 3x² - 5x + 15 and the line equation is y = 2x - 15.

Graphically, you would draw these out and locate their point(s) of intersection.

Algebraically, we attempt to express the pair of simultaneous equations as a single variable equation.

x³ - 3x² - 5x + 15 = 2x - 15

We need to make the right-hand side zero.

x³ - 3x² - 7x + 30 = 0.

We need to make the coefficient of x³ equal two. To do this, we multiply both sides of the equation by 2.

2x³ - 6x² - 14x + 60 = 0

We see that this is of the given form 2x³ + Ax² + Bx + 60 = 0, with A = -6 and B = -14.

Of course, I have no idea what the equations are, so you will need to refer to your question and attempt the question in a similar approach to mine.

So in this question, you are actually not supposed to substitute the value x = 0.6 into that equation given in part (iii).