Respuesta :

we have the equation

[tex]f(x)=x^3-x^2-x+1[/tex]

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

[tex]x^3-x^2-x+1=\left(x-1\right)\left(x^2-1\right)[/tex]

Solve the quadratic equation

[tex]\begin{gathered} x^2-1=0 \\ x^2=1 \\ x=\pm1 \end{gathered}[/tex]

therefore

The equation has 3 real roots

Option C