1. If $x_{1}=1, x_{2}=1-\mathrm{i}, x_{3}=1+\mathrm{i}(\mathrm{i}$ is the imaginary unit $)$ are the three solutions of the equation $x^{3}+a x^{2}+b x+c=0$, then $a+b-c=$ $\qquad$
By Vieta's formulas: $1+1-\mathrm{i}+1+\mathrm{i}=-a \Rightarrow a=-3, 1 \cdot(1-\mathrm{i})(1+\mathrm{i})=-c \Rightarrow c=-2$, $1-\mathrm{i}+1+\mathrm{i}+(1-\mathrm{i})(1+\mathrm{i})=b \Rightarrow b=4 \Rightarrow a+b-c=3$.