Bài giải:

$(ax+b)({{x}^{2}}+cx+1)={{x}^{3}}-3x+2$

Ta có: $(ax+b)({{x}^{2}}+cx+1)=a{{x}^{3}}+ac{{x}^{2}}+ax+b{{x}^{2}}+bcx+b$

= $a{{x}^{3}}+(ac+b){{x}^{2}}+(a+bc)x+b$

Suy ra: $a{{x}^{3}}+(ac+b){{x}^{2}}+(a+bc)x+b$= ${{x}^{3}}-3x+2$

$\Rightarrow \left\{ \begin{align}& a=1 \\ & ac+b=0 \\ & a+bc=-3 \\ & b=2 \\ \end{align} \right.$

$\Rightarrow \left\{ \begin{align}& a=1 \\ & b=2 \\ & c=-2 \\ \end{align} \right.$

Vậy a + b + c = 1 + 2 + (-2) = 1