Let $f(x)=x^3-x^2-x+a$ only have a real negative root then what is range of $a$ 
Let $f(x)=x^3-x^2-x+a$  only have a real negative root then what is range of $a$

I got it : $$g(x)=x^3-x^2-x  \ \ \ h(x)=-a$$
Now we have :
 
so must $-a<-1 \to a>1$

But I want to be algebraic please help me !

Discriminant $\Delta$ of the cubic function
\begin{align} 
f(x)&=ax^3+bx^2+cx+d
\end{align}  
is known to be 
\begin{align} 
\Delta = 18abcd -4b^3d + b^2c^2 - 4ac^3 - 27a^2d^2
,
\end{align}  
and the cubic polynomial has only one real root
when its $\Delta<0$.
For
\begin{align} 
f(x)=x^3-x^2-x+a
\end{align}  
we have 
\begin{align} 
\Delta&=-27a^2+22a+5
,\\
\Delta&<0 \quad\text{when}\quad a\in(-\infty,-\tfrac5{27})
\cup(1,\infty)
\tag{1}\label{1}
.
\end{align}
Let the three roots be 
$x_1=-r$, $x_2=u+v\,i$ and $x_3=u-v\,i$
($r>0$, $u,v\in\mathbb{R}$, $i^2=-1$).
We also know that 
\begin{align} 
-a&=x_1\,x_2\,x_3
,\\
-a&=-r\,(u^2+v^2)
,\\
a&=r\,(u^2+v^2)>0
,
\end{align}
that is, for the real negative root $x_1=-r$
we must have $a>0$.
Hence, together with \eqref{1}, the answer is: $a>1$.
