# Hello everyone. Can anybody please help me with this: x^8 - x^6 + x^4 - x^2 + 1 = 0

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To solve x^8 - x^6 + x^4 - x^2 + 1 = 0.

Let us consider the product (x^8 - x^6 + x^4 - x^2 + 1 )(x^2+1) = (x^8 - x^6 + x^4 - x^2 + 1)+(x^8 - x^6 + x^4 - x^2 + 1)1 = x^10-x^8+x^6-x^4+x^2 +x^8-x^6+x^4-x^2+1.

=> (x^8 - x^6 + x^4 - x^2 + 1 )(x^2+1) = x^10+1.

=> x^8 - x^6 + x^4 - x^2 + 1 = (x^10+1)/(x^2+1) = 0.

=> x^8 - x^6 + x^4 - x^2 + 1 = 0 =>(x^10+1) = 0.

So all the roots of x^10+1 = 0 except the roots of x^2+1 = 0 are the roots of x^8 - x^6 + x^4 - x^2 + 1 = 0.

x^10+1 = 0 => x^10 = -1. So x = (-1/10)^(1/10) . Therefore x^10+1 = 0, or x^10 = -1 has 10 roots complex roots. They are the 10th roots of unity. We use De Moivre's to find the roots:

(cosx+i*sinx)^(1/n) = cos(x/n)+isin(x/n) by De Moivre's theorem.

So x^10 = cos(2n+1)pi+isin2npi), where n is an integer and n > = 0.

We take 10throot of both sides using the De Moivre's theorem:

x = {cos(2n+1)pi+isin(2n+1)/pi)}^(1/10).

x = cos(2n+1)pi/10)+isin(2n+1)pi/10},n is an integer and 0<=n <= 9.

The roots of x^8-x^6+x^4-x^2+1 = 0 are

x = cos(2n+1)pi+isin(2n+1)pi/10 , for n = 0,1, (not 2), 4 , 5,6,( not 7) , 8 and 9 are the 8 roots of x^8 - x^6 + x^4 - x^2 + 1 = 0.

x = cos(2n+1)/10 +isin(2n+1) for n = 2 and n= 7 gives the pair of complex conjugate roots of x^2+1= 0, or the square roots of -1.