Answer
$\dfrac{-27}{4}$
Work Step by Step
Given: $ f(x, y, z)=(x-y)(y-z)=xy-xz-y^2+yz$
The iterated integral can be calculated as:
$\iiint_{\mathcal{B}} f(x,y,z)d V = \iiint_{\mathcal{B}} xy-xz-y^2+yzd V \\
=\int_{0}^{3} \int_{0}^{3} \int_{0}^{1} [xy-xz-y^2+yz] \ dx dy \ dz \\
= \int_{0}^{3} \int_{0}^{3} [\dfrac{x^2y}{2}-\dfrac{x^2z}{2}-xy^2+xyz]_0^1 \ dy \ dz\\=\int_{0}^{3} [\dfrac{y^2}{4}-\dfrac{zy}{2}-\dfrac{y^3}{3}+\dfrac{zy^2}{2}]_0^3 \ dz \\=\int_0^3 (3z-\dfrac{27}{4}) \ dz \\=[\dfrac{3z^2}{2}-\dfrac{27z}{4}]_0^3\\=[\dfrac{3(3)^2}{2}-\dfrac{27(3)}{4}]\\=\dfrac{-27}{4}$
