If $x^2+x+1$ is a factor of $a{x}^3+b{x}^2+cx+d$ (where $a$ and $d$ are equal in magnitude but opposite in sign) then the real root of $a{x}^3+b{x}^2+cx+d=0$ is

Since $x^2+x+1$ is a factor of $a{x}^3+b{x}^2+cx+d$  
$\therefore (a{x}^3+b{x}^2+cx+d)=a(x^2+x+1)(x-\alpha )$ 
Where $\alpha$ is a real root of the given equation.
Equating constant term on both sides, $d=-a\alpha \therefore \alpha =-\frac{d}{a}.$