Application of Triple Integrals Any easy way to solve this integration?
$\int_{0}^{3} \int_{0}^{\frac{6-2x}{3}} \int_{0}^{\frac{6-3y-2x}{2}} xyz \ dzdydx$
My work
$\int \int \int_D xyz \ dV = \dfrac{1}{2}\int_{0}^{3} \int_{0}^{\frac{6-2x}{3}}  xy(z^2 )\biggr|_{0}^{\frac{6-3y-2x}{2}} \ dy \ dx$
$\int \int \int_D xyz \ dV = \dfrac{1}{8}\int_{0}^{3} \int_{0}^{\dfrac{6-2x}{3}}  xy(6-3y-2x)^2 \ dy \ dx$
$\int \int \int_D xyz \ dV = \dfrac{1}{8}\int_{0}^{3} x \left [y\dfrac{(6-3y-2x)^3}{3(-3)} -\dfrac{(6-3y-2x)^4}{4(-3)3(-3)} \right ]\biggr|_{0}^{\dfrac{6-2x}{3}} \ dx$
$\int \int \int_D xyz \ dV = \dfrac{1}{8}\int_{0}^{3} x \left [y\dfrac{(6-3y-2x)^3}{3(-3)} -\dfrac{(6-3y-2x)^4}{4(-3)3(-3)} \right ]\biggr|_{0}^{\dfrac{6-2x}{3}} \ dx$
The terms are expanding, any way to find a solution in easy case. please help

The domain of integration is a tetrahedron with vertices at $(0,0,0),(3,0,0),(0,2,0),(0,0,3)$. The integrand is symmetric with respect to permuting variables. Thus we are allowed to rotate the domain of integration so that slices perpendicular to the $z$-axis are right isosceles triangles and the integral is equal to
$$\int_0^2\int_0^{(6-3z)/2}\int_{(6-3z-2y)/2}xyz\,dx\,dy\,dz$$
We move the $z$ as far out as it will go:
$$\int_0^2z\int_0^{(6-3z)/2}\int_{(6-3z-2y)/2}xy\,dx\,dy\,dz$$
Then rewrite to make the slices of right isosceles triangles clearer. $T(k)$ is the triangle with vertices at $(0,0),(k,0),(0,k)$:
$$\int_0^2z\iint_{T((6-3z)/2)}xy\,dA\,dz$$
Now let's work out what the inner integral is before we substitute $(6-3z)/2$. In other words, what is
$$\iint_{T(k)}xy\,dA=\int_0^k\int_0^{x-k}xy\,dy\,dx$$
This evaluates to
$$\int_0^k x[y^2/2]_0^{x-k}\,dx=\frac12\int_0^k x(x-k)^2\,dx=\frac12[x^4/4-(2/3)kx^3+k^2x^2/2]_0^k=k^4/24$$
So our original integral becomes
$$\frac1{24}\int_0^2z\left(\frac{6-3z}2\right)^4\,dz=\frac{81}{24×16}\int_0^2z(2-z)^4\,dz$$
$$=\frac{27}{128}[8z^2-(32/3)z^3+6z^4-(8/5)z^5+z^6/6]_0^2$$
$$=\frac{27}{128}×\frac{32}{15}=\frac9{20}$$
