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.

Notice that the LHS is:
$$\frac{3n \times (3n-1) \times (3n-2)}{3!} = \frac{n \times (3n-1) \times (3n-2)}{2}$$
$$\to \frac{9n^3-9n^2+2n}{2}$$
For the RHS:
$$3 \times \frac{n\times(n-1)\times(n-2)}{3!} + 6n \times \frac{n\times(n-1)}{2!} + \frac{2n^3}{2}$$
$$\Longrightarrow  \frac{n\times(n-1)\times(n-2)}{2} + \frac{6n^2 \times(n-1)}{2} + \frac{2n^3}{2}$$
$$\to \frac{9n^3-9n^2+2n}{2}$$
Hence, RHS=LHS.
