Relationship between the coefficients of a polynomial equation and its roots My question is-
Solve: $x^3 - 6x^2 + 3x + 10= 0$  given that the roots are in arithmetic progression.

Any help would be greatly appreciated.
Let the three roots be $a-d$, $a$, and $a+d$.

The sum of the roots is the negative of the coefficient of $x^2$, so $(a-d)+a+(a+d)=6$, and therefore $a=2$.

The product of the roots is the negative of the constant term, and therefore $(a-d)(a)(a+d)=-10$. Since we know $a$, we can now find $d$.  We get $a^2-d^2=-5$, and therefore $d^2=9$. 

Remarks: $1.$ In general, if we have a cubic equation $px^3+qx^2+rx+s=0$, where $p\ne 0$, and the roots are $\alpha$, $\beta$, and $\gamma$, then 
$$\alpha+\beta+\gamma=-\frac{q}{p}, \qquad \alpha\beta+\beta\gamma+\gamma\alpha=\frac{r}{p},\qquad \alpha\beta\gamma=-\frac{s}{p}.\tag{$1$}$$
There are analogous relations between coefficients and roots for polynomials of any degree.

$2$. Note that we called the roots $a-d, a, a+d$. We could have called them $a$, $a+d$, and $a+2d$, but the algebra would be a little more complicated. Symmetry is your friend.

$3.$ If the fact we used about roots is not available, there are a couple of things we can do. We can hope that the roots are rational. then they must all divide the constant term $10$. the only candidates are $\pm 1$, $\pm 2$, $\pm 5$, and $\pm 10$, and we quickly get to an answer. 

Or else we can let the roots be $\alpha$, $\beta$, and $\gamma$, and see that the polynomial must be $(x-\alpha)(x-\beta)(x-\gamma)$. Expanding, we find in essence the relationship $(1)$. 

Or else (but this is ugly) we can substitute $a-d$, $a$, and $a+d$ in our polynomial, and expand. Then solve for $a$ and $d$. The equations we get have a fair bit of symmetry, so this is not as difficult as it seems.
First of all the easiest mtd is 

$1)~$ Get one of the roots by hit and trial mtd in your question clearly $~-1~$ is one of the roots i.e when $~x =-1~$ the whole equation results $~0~$ therefore 
$~X=-1 ,~~~ X+1=0~$

$2)~$ apply division method to reduce it to quadratic and then $~u~$ can easily have your roots that are in A. P