If one of the roots of x^5=1 is r and r is not equal to 1, compute (r-1)(r^2-1)(r^3-1)(r^4-1).?

Question

If one of the roots of x^5=1 is r and r is not equal to 1, find (r-1)(r^2-1)(r^3-1)(r^4-1).

Answer 1

Since r is a root of x^5=1,

so r^5=1......(1)



Since r not equal to 1,so r-1 not equal to 0.



So (r^5-1)/(r-1)=0 [From (1)]



=>1+r+r²+r³+r^4=0 ..........(2)

[ Since we have the result

(rⁿ-1)/(r-1)



=1+r+r²+r³+......+r^(n-1) ]



Therefore

( r - 1 ) ( r² - 1 ) ( r³ - 1 ) ( r^4 - 1 )



= ( r - 1 ) ( r^4 - 1 ) ( r² - 1 ) ( r³ - 1 )



= ( r^5 - r^4 - r + 1 ) ( r^5 - r³ - r² + 1 )



= ( 1 - r^4 - r + 1 ) ( 1 -r³ - r² + 1 ) [As r^5=1]



= ( r^4 + r - 2 ) (r³ + r² - 2 )



= [ - ( 1 + r³ + r² ) - 2 ] ( r³ + r² - 2 )



= - ( r³ + r² + 3 ) ( r³ + r² - 2 )



= - [ ( r³ + r² )² + r³ + r² - 6 ]



= - [ r^6 + 2 r^5 + r^4 + r³ + r² - 6 ]



= - [ r + 2 + r^4 + r³ + r² - 6 ]

[Since r^6 = r . r^5 = r . 1 = r ]



= - [ r^4 + r³ + r² + r - 4 ]



= - ( - 1 - 4 ) [ From (2) , r^4 + r³ + r² + r = - 1 ]



= 5

Answer 2

let x=cisΘ

then

(cisΘ)^5=1=cis(n360) wher n is an integer

=> cis5Θ=cis(360n)

=> 5Θ=360n

=> Θ=72n for(n=0,1,2,3,4)(from n=5 we get repeated roots)

(r-1)(r^2-1)(r^3-1)(r^4-1)

=[(r-1)(r^4-1)][(r^2-1)(r^3-1)]

=[r^5-r^4-r+1][r^5-r^2-r^3+1]

=[2-r^4-r][2-r^2-r^3]

=(4-2r^2-2r^3-2r^4+r^6+r^7-2r+r^3+r^4)

=(4-r-r^2-r^3-r^4)



substitute r =cis72



r+r^2+r^3+r^4

=cis72+cis144+cis216+cis288

=cis72+cis144+cis(360-144)+cis(360-72)

= 2Re(cis72)+2Re(cis144)

= 2(cos72+cos144)

=2(-1/2)

=-1



=> (r-1)(r^2-1)(r^3-1)(r^4-1)=

=>(4-r-r^2-r^3-r^4)=4+1=5

Answer 3

A.First solution.



let be :

1,r1,r2,r3,r4 roots of x^5-1=0

this five numbers have a group structure.

r1*r4=1 (are inverse)

r2*r3=1 (inverse)

r1*r1=r2

r2*r2=r4

r3*r3=r1

r4*r4=r3

Sum of roots is zero then

r1+r2+r3+r4+1=0

From here :

prove for r1 (for any of r1,r2,r3,r4 is the same)

(r1-1)(r1^2-1)(r1^3-1)(r1^4-1)=

(r1-1)(r2-1)(r3-1)(r4-1)=

(r1*r4-r1-r4+1)(r2r3-r2-r3+1)=

(2-r1-r4)(2-r2-r3)=

4-2(r1+r2+r3+r4)+(r1+r4)(r2+r3)=

4+2+r1r2+r1r3+r2r4+r3r4=

6+r3+r4+r1+r2=6-1=5



B.

The Second Solution:

(r-1)(r^2-1)(r^3-1)(r^4-1)=

(r^5-r-r^4+1)(r^5+1-r^2-r^3)=

(2-r-r^4)(2-r^2-r^3)=

4-2(r+r^2+r^3+r^4)+(r+r^4)(r^2+r^3)=

(attention (1+r+r^2+r^3+r4)(r-1)=r^5-1=0

then r+r^2+r^3+r4=-1

4+2+r^3+r^4+r^6+r^7=

6+r^3+r^4+r*r^5+r^2*r^5=

6+r^3+r^4+r+r^2=6-1=5



C. Third solution

Use Moivre (Euler)