<s>[INST] Combinatorial proof of $\binom{3n}{3} =3\binom{n}{3} +6n\binom{n}{2} +n^3$? 
Give a combinatorial proof of the following identity: $$\binom{3n}{3} =3\binom{n}{3} +6n\binom{n}{2} +n^3.$$

I've been working on this proof for hours, however I'm not able to show LHS = RHS-
I completely understand binomial theorem and few combinatorial proofs but not able to succeed this one.
Help would be appreciated.
 [/INST] Let $A,B,C$ be the $3$ groups of $n$ students each with grades (marks) $A,B,C$, respectively. You need to select $3$ students.
The LHS is simply the combination of $3n$ students chosen $3$ at a time. 
The RHS is to choose $3,2,1$ or $0$ students from $A,B$ and $C$, respectively:
$${n\choose 3}{n\choose 0}{n\choose 0}+{n\choose 2}{n\choose 1}{n\choose 0}+{n\choose 2}{n\choose 0}{n\choose 1}+{n\choose 1}{n\choose 2}{n\choose 0}+{n\choose 1}{n\choose 1}{n\choose 1}+\\
{n\choose 1}{n\choose 0}{n\choose 2}+{n\choose 0}{n\choose 3}{n\choose 0}+{n\choose 0}{n\choose 2}{n\choose 1}+{n\choose 0}{n\choose 1}{n\choose 2}+{n\choose 0}{n\choose 0}{n\choose 3}=\\
{n\choose 3}\cdot 1\cdot 1+{n\choose 2}\cdot n\cdot 1+{n\choose 2}\cdot 1\cdot n+n\cdot {n\choose 2}\cdot 1+n\cdot n\cdot n+\\
n\cdot 1\cdot{n\choose 2}+1\cdot{n\choose 3}\cdot1+1\cdot{n\choose 2}\cdot n+1\cdot n\cdot{n\choose 2}+1\cdot 1\cdot {n\choose 3}=\\
3{n\choose 3}+6n{n\choose 2}+n^3.$$
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