How find this $a^3+b^3+c^3-20(a+3)(b+3)(c+3)=2013$ equation integer solution if $a,b,c\in Z$,and $a\le b\le c$
and such 
$$\begin{cases}
a+b+c=-3\\
a^3+b^3+c^3-20(a+3)(b+3)(c+3)=2013
\end{cases}$$
Find the value  $3a+b+2c=?$
my try
$$a+b+c=-3\Longrightarrow (a+1)+(b+1)+(c+1)=0$$
so
$$(a+1)^3+(b+1)^3+(c+1)^3=3(a+1)(b+1)(c+1)$$
This problem is from a few month ago comption.
then I can't,Thank you 

Let's forget about $a\le b \le c$ for the moment.
$$a+b+c=-3$$
$$\implies (a+b+c)^3 =a^3+b^3+c^3+3(a+b)(b+c)(a+c)=-27$$
Let's substitute $a^3+b^3+c^3=x$, $(a+b)(b+c)(a+c)=y$.Then, we have that
$$x+20y=2013$$
$$x+3y=-27$$
$$\implies y=120$$
Let's sub $a+b=m,b+c=n,a+c=p$.Now we have
$$m+n+p=-6,$$
$$mnp=120$$
Since by an easy argument either all $m,n,p$ are even or only one of them is, and since $mnp=120$, they are all even, or one is a multiple of $8$. So let's sub again! $m=2r,n=2s,p=2t$.So
$$r+s+t=-3$$
$$rst=15$$
So we have $(r,s,t)=(-5,-1,3)$ and permutations, so a solution is $(a,b,c)=(-1,-9,7)$.
The other case
$$8r+s+t=-6$$
$$rst=15$$
So we have $(r,s,t)=(-1,-3,5)$  and $(a,b,c)=(-8,0,5)$.
Remark: I was running out of letters.
