Prove that if $2a^3 + 27c = 9ab,$ then the roots of $x^3 + ax^2 + bx + c = 0$ form an arithmetic sequence. I am not sure how to begin this problem. Can someone help me?
I have a hint: Let $y = x + \frac{a}{3}$ and rewrite $x^3 + ax^2 + bx + c = 0$ in terms of $y.$
How do I do this?

Isolating $c$ in what we are given, and then plugging it into the polynomial, we have, $$x^3+ax^2+bx+c=x^3+ax^2+bx+\frac{1}{27}(9ab-2a^3)=$$
$$=\frac{1}{27}(3x+a)(9x^2+6ax+9b-2a^2),$$
which says $$x_1=-\frac{a}{3}$$ and
$$x_2+x_3=-\frac{6a}{9}=-\frac{2a}{3}=2x_1$$ and since $$x_2-x_1=x_1-x_3,$$ we are done!
