smc

Let $r_1,r_2,$ and $r_3$ be the roots of $g(x)$. Let $r_4$ be the additional root of $f(x)$. Then from Vieta's formulas on the quadratic term of $g(x)$ and the cubic term of $f(x)$, we obtain the following: $$\begin{array}{rl}{r}_1+r_2+r_3& =-a\\ r_1+r_2+r_3+r_4& =-1\end{array}$$ Thus $r_4=a-1$. Now applying Vieta's formulas on the constant term of $g(x)$, the linear term of $g(x)$, and the linear term of $f(x)$, we obtain: $$\begin{array}{rl}{r}_1{r}_2{r}_3& =-10\\ r_1{r}_2+r_2{r}_3+r_3{r}_1& =1\\ r_1{r}_2{r}_3+r_2{r}_3{r}_4+r_3{r}_4{r}_1+r_4{r}_1{r}_2& =-100\end{array}$$ Substituting for $r_1{r}_2{r}_3$ in the bottom equation and factoring the remainder of the expression, we obtain: $$-10+(r_1{r}_2+r_2{r}_3+r_3{r}_1)r_4=-10+r_4=-100$$ It follows that $r_4=-90$. But $r_4=a-1$ so $a=-89$ Now we can factor $f(x)$ in terms of $g(x)$ as $$f(x)=(x-r_4)g(x)=(x+90)g(x)$$ Then $f(1)=91g(1)$ and $$g(1)=1^3-89\cdot 1^2+1+10=-77$$ Hence $f(1)=91\cdot (-77)=\boxed{-7007}$.