複數的數學問題求，急

設X=COS2/7π +iSIN2/7π x^7 = (cos2π/7 + i*sin2π/7)^7= cos2π + i*sin2π ;;;戴美弗定理= cos(2π)= 1
=> 0 = x^7 -1 = (x - 1)(1 + x + x^2 + x^3 + x^4 + x^5 + x^6) => 1 + x + x^2 + x^3 + x^4 + x^5 + x^6 = 0......(1)=> x^9 = x^7*x^2 = x^2......(2)=> x^8 = x, x^10 = x^3, x^11 = x^4......(3)

w = (1+x)(1+x^2)(1+x^3)(1+x^4)(1+x^5)(1+x^6) = (1+x+x^2+x^3)(1+x^3)(1+x^4)(1+x^5)(1+x^6) = (1+x+x^2+2x^3+x^4+x^5+x^6)(1+x^4)(1+x^5)(1+x^6) = x^3(1+x^4)(1+x^5)(1+x^6) ;;; by Eq.(1)= x^3(1+x^4+x^5+x^9)(1+x^6)= x^3(1+x^2+x^4+x^5)(1+x^6) ;;; by Eq.(2)= x^3(1+x^2+x^4+x^5+x^6+x^8+x^10+x^11)= x^3(1+x+x^2+x^3+2x^4+x^5+x^6) ;;; by Eq.(3)= x^3*x^4= x^7= 1

z^7-1=0的7個根為1,x,x^2,x^3,x^4,x^5,x^6
(z^7-1)=(z-1)(z-x)(z-x^2)(z-x^3)(z-x^4)(z-x^5)(z-x^6)
z^6+z^5+z^4+z^3+z^2+z+1=(z-x)(z-x^2)(z-x^3)(z-x^4)(z-x^5)(z-x^6)
代入z=-1即得(1+x)(1+x^2)(1+x^3)(1+x^4)(1+x^5)(1+x^6)=1

x^7=1
x+x^2++x^3+x^4+x^5+x^6=0
(1+x)(1+x^2)(1+x^3)(1+x^4)(1+x^5)(1+x^6)
=(1+x)(1+x^2)(1+x^4)(1+x^3)(1+x^5)(1+x^6)
=(1+x+x^2+x^3)(1+x^4)(1+x^3+x^5+x^8)(1+x^6)
=(1+x+x^2+x^3+x^4+x^5+x^6+x^7)(1+x^3+x^5+x^8+x^6+x^9+x^11+x^14)
=(1+x+x^2+x^3+x^4+x^5+x^6+x^7)(1+x^3+x^5+x+x^6+x^2+x^4+1)
=(1+0)*(1+0)
=1