By the Vieta's formulas we have −ab=(α−β)+α+(α+β)=3α, ac=(α−β)α+(α−β)(α+β)+α(α+β)=
=α2−αβ+α2−β2+α2+αβ=3α2−β2, −ad=(α−β)α(α+β)=α3−αβ2
Then (2b2−9ac)b+27a2d=a3(2(ab)2−9ac)ab+27a3ad
So we need to prove that (2(ab)2−9ac)ab+27ad=0
We have 2(ab)2−9ac=2(3α)2−9(3α2−β2)=−9α2+9β2, so (2(ab)2−9ac)ab+27ad=
=(−9α2+9β2)(−3α)+27(−α3+αβ2)=0