Given 1 + X + x square +... + x2004 + x2005 = 0, find the value of x2006

Given 1 + X + x square +... + x2004 + x2005 = 0, find the value of x2006

From 1 + X + x square +... + x2004 square + x2005 square = (1-x2006 Square) / (1-x) = 0 (this is the summation formula of equal ratio sequence), so 1-x2006 square = 0 and X ≠ 1, that is, x = - 1, so 1-x2006 square = 1

Given x + 1 / x2 = 2, find the value of x2 + 1 / X2, where X2 is the square of X

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

Solve an equation x (1 power) + x2 + X3 = 5 x2 + X3 + X4 = 1 X3 + X4 + X5 = - 5 X4 + X5 + X1 = - 3 X5 + X1 + x2 = 2 to find the value of X,

5,
3（x1+x2+x3+x4+x5）=1+5-5-3+2=0
So X1 + x2 + X3 + X4 + X5 = 0
X1 + x2 + X3 = 5, X4 + X5 + X1 = - 3, the two formulas add: X1 + (x1 + x2 + X3 + X4 + x5) = 2, X1 = 2
Similarly, X2 = 3, X3 = 0, X4 = - 2, X5 = - 3