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To solve x^4+x^3+x^2+x+1 = 0.
We know that 1+x+x^2+x^4 = (x^5-1)/(x-1).
So the given equation is therefore rewritten as:
(x^5-1) /(x-1) = 0
Multiply by (x-1). So , mind x-1 is our factor.
(x^5 -1) = 0
x^5 = 1.
We know 1 = cos2npi +isin2npi.
x^5 = cos2npi+isin2npi
Take the 5 th root.
x = (co2npi+isin2npi) ^(1/5)
x = (cos(2npi)/5 +isin(2npi)/5 , for n = 0,1,2,3,4....by D'Moivres theorem.Actually after n=4, for the next integral values the roots repeat.
x0 = 1 is not the root as we have multiplied by our factor (x-1) to the given expression (x^4+x^2+x^3+1 ).
x1 = cos72+isin72
x2 = cos144 + isin 144
x3 = cos216 +i sin 216
x4 = cos288 +isin288.
So x1,x2,x3 and x4 are the solutions.
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