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?
 A: Well, do exactly what the hint tells you!
$$(y - \frac{a}{3})^3 + a(y-\frac{a}{3})^2 + b(y-\frac{a}{3}) + c = 0$$ 
$$\frac{1}{27}(2a^3 - 9a^2y - 9ab + 27by +27c + 27y^3)$$
$$\frac{1}{27}( - 9a^2y + 27by + 27y^3)$$
$$\frac{y}{27}(27y^2 - 9a^2 + 27b)$$
Now can you see what the roots are? How does this relate to the roots of your original polynomial?
A: The roots satisfy 
$$x_1+x_2+x_3=-a\tag 1$$
Write the given cubic equation in its depressed form with $t=x+\frac a3$,
$$t^3+(b-\frac {a^3}3)t+\frac{2a^2+27c-9ab}{27}=0$$
which, from the given condition, reduces to,
$$t^3+(b-\frac {a^3}3)t=0$$
and one of the roots is $t_1=0$, or, $x_1=-\frac a3$. From (1), we have 
$$x_2+x_3=-a-x_1=-\frac23a=2x_1$$
hence, an arithmetic sequence.
A: 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!
A: WLOG let us assume the roots to be $p-q,p,p+q$
$$-a=p+q+p+p-q\iff p=-\frac a3$$
$$c=p(p^2-q^2)=-\frac{a}3\left(\frac{a^2}{9}-q^2\right)\implies q^2=?$$
$$b=p(p+q)+(p-q)(p+q)+p(p-q)=p^2-q^2+2p^2=3\left(-\frac{a}3\right)^2-q^2\implies q^2=?$$
Try to compare the two values of $q^2$ to eliminate $q$. 
A: Suppose that $u,v,w$ are the three roots of $x^3+ax^2+bx+c$. Then, because
$$
(x-u)(x-v)(x-w)=x^3-(\overbrace{u+v+w}^{-a})x^2+(\overbrace{uv+vw+wu}^b)x-\overbrace{\ \ u\ v\ w\ \ }^{-c}
$$
we have
$$
2(\overbrace{u+v+w}^{-a})^3+27\overbrace{\ \ u\ \ v\ \ w\ \ }^{-c}-9(\overbrace{u+v+w}^{-a})(\overbrace{uv+vw+wu}^b)\\
=\underbrace{(u+v-2w)}_{\substack{\text{$0$ if $w$ is midway}\\\text{between $u$ and $v$}}}\underbrace{(2u-v-w)}_{\substack{\text{$0$ if $u$ is midway}\\\text{between $v$ and $v$}}}\underbrace{(u-2v+w)}_{\substack{\text{$0$ if $v$ is midway}\\\text{between $u$ and $w$}}}
$$
