My title is very poorly worded, I apologise, I'm having a hard time wording the question. My lecturer described it as being obvious, so I must be missing something very fundamental (the substitution is in terms of $a_2$, not $\alpha _2$.)
Thanks for any help!
Upon substituting we get $(2 a_2^3)/27 - (a_2 a_1)/3 + a_0 - (a_2^2 y)/3 + a_1 y + y^3$. When I substitute $y = x + a_2 / 3$ back in it, wolframalpha gives the original polynomial.
I can't see how we'd ever get anything different.
One way to see it is the binomial expansion $(x+y)^3=x^3+3x^2y+3xy^2+y^3$, you now need the $x^2$ terms to cancel out (the rest of the factors contribute only lower power terms) so setting $3y=a_2$ does the trick.