**Question:**

Let (C \subset \mathbb{ZxZ} ) be the set of integer pairs ((a,b)) for which the three complex roots (r_1,r_2,r_3) of the polynomial (p(x)=x^3-2x^2+ax-b) satisfy (r_1^3+r_2^3+r_3^3=0). Then the cardinality of (C) is

A. (\infty)

B. 0

C. 1

D. (1<|C|<\infty)

**Discussion:**

We have (r_1+r_2+r_3=-(-2)=2)

(r_1r_2+r_2r_3+r_3r_1=a) and

(r_1r_2r_3=-(-b)=b)

Also, of the top of our head, we can think of one identity involving the quantities (r_1^3+r_2^3+r_3^3=0) and the three mentioned just above.

Let’s apply that:

(r_1^3+r_2^3+r_3^3-3r_1r_2r_3=(r_1+r_2+r_3)(r_1^2+r_2^2+r_3^2-(r_1r_2+r_2r_3+r_3r_1))) …(…(1))

Also, note that (r_1^2+r_2^2+r_3^2=(r_1+r_2+r_3)^2-2(r_1r_2+r_2r_3+r_3r_1) )

So (r_1^2+r_2^2+r_3^2=(2)^2-2(a)=4-2a ).

So by equation ((1)),

(0-3b=2(4-2a-a))

or, (6a-3b=8).

Now, we notice the strangest thing about the above equation: (3|(6a-3b)) but (3) does not divide (8).

So, we get a contradiction.

Why did this contradiction occur? We didn’t start off by saying “assume that … something …”. Well, even though we pretend to not assume anything, we did assume that this equation has a solution for some (a,b). So, the

ANSWER: (|C|=0).