Show that the roots of x^{5 }=1 can be written as 1, `alpha^1,alpha^2,alpha^3,alpha^4` .hence show that `(1-alpha^1) (1-alpha^2) (1-alpha^3)(1-alpha^4)=5.`

#### Solution

`x^5=1=cos0+i sin0`

`thereforex^5=cos(2kpi)+i sin(2kpi)`

`therefore x^1=(cos(2kpi)+i sin(2kpi))^(1/5)=cos((2kpi)/5)+i sin ((2kpi)/5)`

Putting k=0,1,2,3,4 we get the five roots as

`x_0=cos0+i sin0=1`,

`x_1=cos ((2pi)/5)+isin((2pi)/5)`,

`x_2=cos((4pi)/5)+isin((4pi)/5)`,

`x_3=cos((6pi)/5)+isin((6pi)/5),`

`x_4=cos((8pi)/5)+isin((8pi)/5).`

Putting `x_1=cos((2pi)/5)+isin((2pi)/5)=alpha` we see that `x_2=alpha^2, x_3=alpha^3, x_4=alpha^4`

∴ the roots are 1, `alpha, alpha^2, alpha^3, alpha^4` and hence

`therefore x^5-1=(x-1)(x-alpha)(x-alpha^2)(x-alpha^3)(x-alpha^4)`

`therefore (x^5-1)/(x-1)=(x-alpha)(x-alpha^2)(x-alpha^3)(x-alpha^4)`

`therefore (x-alpha)(x-alpha^2)(x-alpha^3)(x-alpha^4)=x^4+x^3+x^2+x^1+1.`

Putting x = 1, we get

`(1-alpha)(1-alpha^2)(1-alpha^3)(1-alpha^4)=5`