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.

Brute force:
$27n^{3} - 27n^{2}+ 6n = 27n^{3} - 27n^{2} + 6n$
$27n^{3} - 9n^{2} - 18n^{2} + 6n = 3n^{3} - 3n^{2} - 6n^{2} + 6n + 18n^{3} - 18n^{2} + 6n^{3}$
$(9n^{2} - 3n)(3n - 2) = (3n^{2} - 3n)(n - 2)  + 18n^{2}(n-1) + 6n^{3}$
$3n(3n - 1)(3n - 2) = 3n(n - 1)(n - 2)  + 3(6n)n(n-1) + 6n^{3}$
$3n(3n - 1)(3n - 2)/6 = 3n(n - 1)(n - 2)/6  + (6n)n(n-1)/2 + n^{3}$
$\binom{3n}{3} =3\binom{n}{3} +6n\binom{n}{2} +n^3$
