x4+x3+x2+x+1=0 (4,3,2-->exponents) Could anyone please solve this without D'Moivres theorem???

x^4+x^3+x^2+x+1 = 0.

Divide
by x^2.

x^2+x+1+1/x+1/x^2 =
0.

(x^2+1/x^2) +(x+1/x) +1 =
0

(x+1/x)^2 -2 + (x+1/x)+1
=0

(x+1/x)^2 +(x+1/x) -1 = 0. This is aquadratic equation
in (x+1/x). Or

y^2 +y -1 =  0 , where y =
x+1/x.

y1 = (-1+sqrt5)/2 , or y2 =
(-1-sqrt5)/2

y1 = x+1/x = 
(-1+sqrt5)/2

x^2 +(1-sqrt5)/2 *x  +1 =
0

x1  =  {(-1+sqrt5)/2 + sqrt[(1-sqrt5)^2/4
-4}/2

x1 = {-(1-sqrt5)/4 +
sqrt(-5-sqrt5)/2}/2

x1 = {-(1-sqrt5)/4
+sqrt(-1)sqrt(5+sqrt5)/4}

x2 = {-(1-sqrt5)/4 -sqrt(-1)
sqrt(5+sqrt5)/4

Similarly we can solve for y2 = x+1/x =
(-1-srqt5)/2

x^2 +(1+sqrt5)x+1 =
0

x3 = -(1+sqrt5) /4+ [sqrt(sqrt5 -5)]/4
or

x3  = -(1+sqrt5)/4
+{sqrt(-1)sqrt(5-sqrt5)}/4

x4 = -(1+sqrt5)- {sqrt(-1)
sqrt(5-sqrt5)}/4

S0 there are as above 4 complex
roots x1,x,2 ,x3 and x4..