Verified answer

Don't overthink it. You want to put an "a²" wherever you have a "2x - 1", so

2x - 1 = a²

2x = a² + 1

x = (a² + 1)/2

Now plug this expression into your f(2x -1):

8[(a² + 1)/2]^3 - 12[(a² + 1)/2]² + 4[(a² + 1)/2]

[8(a² + 1)^3]/8 - 12(a² + 1)²/4 + 4(a² + 1)/2

(a² + 1)^3 - 3(a² + 1)² + 2(a² + 1)

a^6 + 3a^4 + 3a² + 1 - 3a^4 - 6a² - 3 + 2a² + 2

a^6 - a²

[In this form, it may be easier to see that f(x) = x^3 - x. You can check your answer by plugging in (2x - 1). It works!]