Rearranging equations with alpha and beta

ab = 1      a + b = -4  →   b = -4 - a

Subbing the last equation into the first, we have

a[-4 - a]  = 1      simplify

-4a - a^2  = 1     rearrange

a^2 + 4a + 1  = 0      using the quadratic formula to solve, we have

a = √3 - 2      or    a =    -√3 - 2

Which means that b =    -4 - [ √3 - 2] =  -2 - √3     or   b =  -4 - [  -√3 - 2  ] =  √3 - 2

So

(a^2  - b) * (b^2 - a)  =  [ ( √3 - 2)^2  - ( -2 - √3) ] * [ ( -2 - √3)^2  - (√3 - 2) ]  =

[ 3 -4√3 + 4 + 2 + √3] *  [ 4 + 4√3 + 3 - √3 + 2]  =

[ 9 - 3√3] * [ 9 + 3√3 ]  =

81 - 27 = 

  =  54

Evaluating this when a =  -√3 - 2   and b =  √3 - 2    gives exactly the same result  as shown below

(a^2  - b) * (b^2 - a)  =  [ ( -√3 - 2)^2  - ( √3 - 2) ] * [ ( √3 - 2 )^2  - (-√3 - 2) ]  = 54

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 + b^{2  =}  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