Find coefficient of $x^3y^4z^5$ in polynomial $(x + y + z)^8(x + y + z + 1)^8$

Objective: To find coefficient of $\ x^3y^4z^5\ $ in $\ (x + y + z)^8(x + y + z + 1)^8$

First expand $$(1+(x+y+z))^8=\sum_{r=0}^{8} {8 \choose r}(x+y+z)^r$$ Now multiplying $(1+(x+y+z))^8$ with $(x+y+z)^8$ gives us,

$$(x+y+z)^8(1+(x+y+z))^8=\sum_{r=0}^{8} {8 \choose r}(x+y+z)^{r+8} \label{1} \tag{1}$$

Now using multinomial theorem, coefficient of $x^3y^4z^5$ in $(x+y+z)^n$ is $\Large{}{n \choose {3,\ 4,\ 5}}$ only when $n=12$.

Applying the same logic on equation $(1)$, we get the coefficient as

$$\sum_{r=0}^{8} {8 \choose r}* {8+r \choose {3,\ 4,\ 5}}$$ so we can fix $r=4$ and the answer we get is $${8\choose4}*{12\choose{3,4,5}}$$

For computationally rigorous answer see Wolfram Alpha coefficient finder