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 !
 A: Hint:
By Descartes rule of signs, we have that a polynomial $f(x)$ cannot have more negative roots than the number of changes of sign in $f(-x)$.
Note that $f(-x)=-x^3-x^2+x+a$. So, ??
A: HINT
Find the minimum of the function and set the value of $a$ such that it is greater than 0.
A: 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: $a>1$.
A: $f'(x)=(3x+1)(x-1)$. By checking the sign of $f''$, you can say that $f$ has a local maxima at $x=-1/3$ and a local minima at $x=1$. Now as $f$ doesn't have any positive root, one should have $f(1)>0$, which implies $a>1$.
A: Let $b$ be the only real and negative root. Then $b=-c$ for some $c>0$ and we have $$a-1=c^3-c^2+c -1= (c-1)^2(c+1)\geq 0$$
So $a\geq 1$. If $a=1$ we get two $c$ namely $1$ and $-1$ so the given equation has 2 real soltion so $a>1$.
A: $f(x)=x^3-x^2-x+a$
$f'(x)=3x^2-2x-1=(x-1)(3x+1)$
$\begin{array}{|c|ccccc|}\hline
x & -\infty && -\frac 13 &  & 1 && +\infty \\\hline
f'&& + && - && +\\\hline
f &-\infty &\nearrow & a+\frac 5{27} &\searrow & a-1 & \nearrow & +\infty\\\hline
\end{array}$
Since $f$ is continuous it has a root in $]-\infty,-\frac 13]$ when $a>-\frac 5{27}$ by applying intermediate value theorem.
If we want this root to be the only one (with $x<0$) then the local minimum in $x=1$ should be positive so $a>1$.
$a>1\implies a>-\frac 5{27}$ thus $a>1$ is solution of the problem.
