Let me try to solve this in a different way ...
ab = 1 a + b = -4
(a2 - b) * (b2 - a)
= a2b2 - a3 - b3 + ab
Since ab = 1 ---> a2b2 = (ab)2 = (1)2 = 1
So: [a2b2] - a3 - b3 + [ab] = [ 1 ] - a3 - b3 + [1] = 2 - a3 - b3 = 2 - [ a3 + b3 ]
But: a3 + b3 can be factored as (a + b)(a2 - ab + b2)
which equals: -4(a2 - ab + b2)
So 2 - [ a3 + b3 ] = 2 - -4(a2 - ab + b2) = 2 + 4(a2 - ab + b2)
Now, let's look at a2 - ab + b2: -ab can be written as 2ab - 3ab,
so: a2 - ab + b2 = a2 + 2ab + b2 - 3ab
Since a2 + 2ab + b2 can be factored as (a + b)2
therefore: a2 - ab + b2 = a2 + 2ab + b2 - 3ab = (a + b)2 - 3ab
Putting this together, we get: 2 + 4(a2 - ab + b2) = 2 + 4[ (a + b)2 - 3ab ]
Substituting -4 for a + b and 1 for ab:
2 + 4[ (a + b)2 - 3ab ] = 2 + 4[ (-4)2 - 3(1) ] = 2 + 4[ 16 - 3 ] = 2 + 4[13] = 2 + 52 = 54