If $\frac{1+i\sqrt{3}}{2}$  is a root of the equation $x^4-x^3+x-1=0$ , then its real roots are

Substitute 1 in the above equation to check whether it is a root

${\left(1\right)}^4-{\left(1\right)}^3+1-1$

$1-1+1-1=0$

So 1 is a root

Substitute $-1$  in the equation

${\left(-1\right)}^4-{\left(-1\right)}^3+\left(-1\right)-1$

$1-1+1-1=0$

So $-1$  is a root

So $1,-1$  are the roots