Is there a way to expand the cube of a trinomial easily? 
Is there a way to expand $(x+y+z)^3$ easily/fast?

It takes some time for me to expand this type of quantities, I was just thinking if there's like a shortcut. 
Motivation:  whenever I do $u$-substitutions in my integral calculus, my professor always wants me to expand it and I'm having problems when the final answer is like $\frac{1}{3} (x^3+2x+3)^3$, it consumes my time.
 A: In $(x+y+z)^n$, the coefficient of $x^ay^bz^c$, where $a+b+c=n$, is $\frac{n!}{a!b!c!}$.
Explain: For each $x^ay^bz^c$ you choose a of "x", b of "y", c of "z" in each of $(x+y+z)$, therefore you have $C^n_aC^{n-a}_bC^{n-a-b}_c=\frac{n!}{a!b!c!}$ many ways to do it.
A: You can use $(a+b+c)^3=\sum\limits_{cyc}(a^3+3a^2b+3a^2c+2abc)$.
A: If $z=0$ you get $x^3+3x^2y+3xy^2+y^3$. All terms will have degree $3$ so all terms which contain only two of $x, y,z$ will be of the form $a^3$ or $3ab^2$ so the terms including $z$ can be written down (all forms appear by symmetry).
Then there is one term $xyz$ which  is not of this form. You can take any letter from the first bracket, one of two from the second and the third is forced. This gives a coefficient of $6$.
If expanding $(a+b+c+d)^3$ note that terms have at most three letters, so the forms of the terms and the coefficients can be determined by expanding $(a+b+c)^3$ and then adding the symmetrically equivalent terms involving $d$.
A: This identity is fairly well known, and straight-forward to verify from sight:
$(a+b+c)^3 = a^3 + b^3 + c^3 + 3(a+b+c)(ab+bc+ac) - 3abc$
Note that the term count nets out correctly as 3 + 27 - 3 = 27; and the cardinality of the terms originating from the complex product are correct.
