$x^3-6x^2+11x+m=0$ roots in arithmetic progression 
Given equation:
  $$x^3-6x^2+11x+m=0.$$
  For which values of $m$ roots of the equation are  roots in arithmetic progression?

I've found a version of this exercise in which the equation could be converted to a quadratic by letting $y$ be $x^4$, but I couldn't really come out with a method that works here as well from that.
I've applied Vieta's and got $x_1^3+x_1^23r+2x_1r^2+m=0$(from $x_1*x_2*x_3=-m$, r being the ratio of the progression) and also $x_1= \frac{11-2r^2}{6+3r}$(from working with Vieta's) but I don't think replacing $x_1$ in the previous one would be the way to go. 
Can I have some hints on how to approach this? Thank you.
 A: Let $x_1$, $x_2$ and $x_3$ be our roots.
Thus, $$x_1+x_2+x_3=6$$ or
$$3x_2=6,$$ which gives $x_2=2$ and from here we can get the value of $m$:
$$8-24+22+m=0$$ or
$$m=-6$$ and easy to see that this value is valid:
$$x^3-6x^2+11x-6=(x-1)(x-2)(x-3).$$
Done!
A: Using the roots $a-d,a,a+d$, the conditions are
$$3a=6,\\
3a^2-d^2=11,\\
a^3-ad^2=-m.$$
Then
$$a=2,d=\pm1,\\m=-6$$ and this is the only solution (though there are two progressions, $1,2,3$ and $3,2,1$).
A: If we deflate the polynomial using $y=x-2$, the roots remain in an arithmetic progression
$$x^3-6x^2+11x+m\to y^3-y+m+6.$$
The deflated polynomial needs to be odd (of the form $y(y^2-d^2)$), and this works with
$$m+6=0.$$
A: HINT:
The roots of $P$ ($\deg 3$) are in arithmetic progression if and only if their arithmetic mean $\frac{s}{3}$ is a root, that is 
$$P(\frac{s}{3})=0$$
A: you will have $$x_1+x_2+x_3=6$$ and $$x_1x_2+x_1x_3+x_2x_3=11$$ and $$x_1x_2x_3=-m$$ where
$$x_2=x_1+d,x_2=x_1+2d$$
A: Let the roots be $a,a+d,a+2d$ then we have 
\begin{eqnarray*}
a+a+d+a+2d=3(a+d)=6 \\
\color{red}{a+d=2} \\
a(a+d)+a(a+2d)+(a+d)(a+2d)=\color{blue}{3a^2+6ad+2d^2 =11}
\end{eqnarray*}
Now square $\color{red}{a+d=2}$ , multiply it by $3$ and subtract the last equation to get $d^2=1$. It is easy form here ?
A: Let $P(x)=x^3-6x^2+11x+m$.
Change $x=y+2$ to get
$$Q(y)=P(x+2)=(y+2)^3-6(y+2)^2+11(y+2)+m=y^3-y+6+m$$
The roots of $P$ are in arithmetic progression if and only if the roots of $Q$ either are.
Since the sum of the roots of $Q$ are $0$, the roots of $Q$ are $-u,0,u$ for some $u\ge 0$. Since $0$ is a root of $Q$, we get $6+m=0$.
