we have the equation
![f(x)=x^3-x^2-x+1](https://img.qammunity.org/2023/formulas/mathematics/college/ymzvw6c37vitkdfol45wrkfz29dolr9xlc.png)
For x=1
f(x)=0
that means------> x=1 is a real root
Divide the given function by the factor (x-1)
x^3-x^2-x+1 : (x-1)
x^2-1
-x^3+x^2
-------------------
-x+1
x-1
---------
0
therefore
![x^3-x^2-x+1=\left(x-1\right)\left(x^2-1\right)](https://img.qammunity.org/2023/formulas/mathematics/college/oxyukkw80gobmib22ymbxz5tub5z5ysmfd.png)
Solve the quadratic equation
![\begin{gathered} x^2-1=0 \\ x^2=1 \\ x=\pm1 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/4vwpdiopq8yjxhwukw24174zc02p58bqap.png)
therefore
The equation has 3 real roots
Option C