If $x^2+x+1$ is a factor of $f (x)= ax^3+bx^2+cx+d$ , then the real root of $f (x)=0$ is If $x^2+x+1$  is a factor of  $f (x)= ax^3+bx^2+cx+d$  , then the real root of $f (x)$ is  
A) $-d/a$
B) $d/a$
C) $a/d $
D) none of these 
My try 
I take $f (x) = ( x^2+x+1)(x+1)  = x^3+2x^2+2x+1$
Real root $x=  -d/a =-1$
And also  
$f (x) = ( x^2+x+1)(x-2)  = x^3-x^2-x-2$
Real root $x =  -d/a =2$
But how can I prove it in general? 

If you divide $f (x)= ax^3+bx^2+cx+d$ to $x^2+x+1$ : you will have 
$$f (x)= ax^3+bx^2+cx+d=(x^2+x+1)(ax+(b-a))+x(a+c-b+a)+(a-b+d)$$ when remainder is $0  $ so $$ax+(b-a)=0 \to x=\frac{a-b}{a} \tag{1}\\
\forall x:\begin{cases}a+c-b+a=0 \\a-b+d=0\end{cases}$$ now from 3rd equation we have $d=-(a-b)$ put into (1)
$$x=\frac{a-b}{a}=x=\frac{-d}{a}$$
