Find the sum of $x_1+x_2+x_3$ of intercept points Suppose that the straight line $L$ meets the curve $y=3x^3-15x^2+7x-8$ in three points $(x_1,y_1)$, $(x_2,y_2)$ and $(x_3,y_3)$. Then $x_1+x_2+x_3=?$
A) 3  $\quad$   B) 4   $\quad$       C) 5   $\quad$    D) 6  $\quad$      E) 7
At the beginning, my main idea is to use Vieta's theorem $x_1+x_2+x_3=-\frac{b}{a}$, and find the answer is $5$, it is correct if $y=0$. But when I use software to draw the grapic of $y=3x^3-15x^2+7x-8$ and a line intercept at three points,I also changed the coefficient of the straight line and use calculator did the summation,still can get $x_1+x_2+x_3=5$, I want to know the general solution of this question  

Let $y=ax+b$ the equation of the straight line $L$. Then $x_1,x_2,x_3$ are solutions of the equation
$3x^3-15x^2+(7-a)x-8-b=0$.
Then Vieta says: $x_1+x_2+x_3= - \frac{-15}{3}=5.$
