Compute $(1 + i)(1 + 2i)(1 + 3i)$.
Expanding, gives us \(\left(1+i\right)\left(1+2i\right)=-1+3i\). Then, expanding again gives us \(\left(-1+3i\right)\left(1+3i\right)=\boxed{-10}.\)
(1 + i)(1 + 2i)(1 + 3i)
= (1 + 2i - i) (1 + 2i + i) (1 + 2i)
= ((1 + 2i)^2 + 1)(1 + 2i)
= (-3 + 4i + 1)(1 + 2i)
= (-2 + 4i)(1 + 2i)
= -2(1 - 2i)(1 + 2i)
= -2 (5)
= -10
