Quadratic: x^2=2x-3 show a^3=a-6

Given that alpha (I use 'a' and 'b' from here for alpha and beta respectively) is a root of x^2=2x-3 show (i) a^3=a-6 and (ii) a^2 - 2a^3=9

My working so far is to rearrange the quadratic to x^2 - 2x +3 = 0
Which is of the form x^2 - (sum of roots)x + (product of roots) = 0
Thus a+b = 2 and a.b = 3

My thought was that the required properties would be shown by substituting these values into the equations such as:
a^2 + b^2 = (a+b)^2 - 3.a.b(a+b) etc
but these appear to resolve to a^2 - 2a + 3 = 0 (a somewhat circular argument).

Where should I be directing my thoughts? Thanks in advance.

You don't really need b in this case. Since a is a root, we have

a^2 = 2a - 3

so we can reduce powers of a. We get
(i) a^3 = a*(a^2)
= a*(2a-3)
= 2a^2 - 3a
= 2(2a-3) - 3a
= 4a - 6 - 3a
= a - 6

(ii) a^2 - 2a^3
= (2a-3) - 2a(2a-3)
= 2a-3 - 4a^2 + 6a
= 8a - 3 - 4(2a-3)
= 8a - 3 + 12 - 8a
= 9


Through a similar process, any polynomial expression in a can be reduced to a linear expression in a, so that equality testing among these expressions can be performed in general. More general still, we don't need to start with a quadratic; an nth degree polynomial suffices, though the final expression will then have degree < n. Testing equality then reduces to figuring out which "polynomial combinations" c0 + c1*a + ... + c_(n-1)*a^(n-1) give 0. This line of thought leads to a beautiful theorem in field extensions, namely that the only combination that gives 0 is the trivial one if and only if the nth degree polynomial was irreducible (over the field in which its coefficients lie).

Since the determinant is 4 - 12 = -8 there are two different complex roots.

Note sqrt(-8) = sqrt(4 * -2) = sqrt(4) * sqrt(-2) = 2i*sqrt(2)

a = [2 +/- 2i*sqrt(2)] / 2 - Divide out the 2s

a = (1 +/- i*sqrt(2))