# Prove that (x^8+x^6+x^4+x^2+1)/(x^4+x^3+x^2+x+1) = x^4-x^3+x^2-x+1.

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### 2 Answers

To prove that (x^8+x^6+x^4+x^2+1)/(x^4+x^3+x^2+x+1) = x^4-x^3+x^2-x+1.

We know that x^8+x^6+x^4+x^2+1 is a GP with x^2 as common ratio..

So x^8+x^6+x^4+x^2+1 = (x^10-1)/(x^2-1)

Similarly, x^4+x^3+x^2+x+1 = (x^5-1)/(x-1).

Therefore (x^8+x^6+x^4+x^2+1)/(x^4+x^3+x^2+x+1) = (x^10-1)/(x^2-1)}/{x^5-1)/(x-1) = (x^10-1)(x-1)/{(x^5-1)(x^2-1}

(x^10 -1) = (x^5+1)(x^5-1).

Therefore (x^10-1)(x-1)/(x^5-1){(x^2-1)} = (x^5+1)(x^5-1)(x-1)/{(x^5-1)(x+1)(x-1)} = (x^5+1)/(x+1).

(x^5 - (-1)^5}/(x- (-1)) = (x^4 +x^3*(-1) +x^2*(-1)^2+x(-1)+(-1)^3+ (-1)^4 = x^2 -x^3+x^2-x+1.

Therefore (x^8+x^6+x^4+x^2+1)/(x^4+x^3+x^2+x+1) = x^4-x^3+x^2-x+1.

We’ll apply the following formula:

a^n – b^n= (a-b)*[a^(n-1) + a^(n-2)*b + ….+b^(n-1)]

We’ve noticed that both, denominator and numerator of the left side ratio, have the second factor from the development above.

We'll re-write the left side:

[( x^10) – 1]/[(x^2)-1]

(x^5)-1 / (x-1)

[( x^10) – 1/(x^2)-1] / [ (x^5)-1 / x-1]= [( x^10) – 1/(x^2)-1]*[( x-1)/( (x^5)-1)]

But ( x^10) – 1=((x^5)^2)-1=(x^5 - 1)(x^5 + 1)

and x^2 – 1=(x-1)(x+1)

We’ll substitute in the found ratio the latest results:

[( x^10) – 1/(x^2)-1]*[( x-1)/( (x^5)-1)]=[(x^5 - 1)(x^5 + 1)/ (x-1)(x+1)]*[( x-1)/( (x^5)-1)]

After simplifying of the similar terms, the result will be:

(x^5 + 1)/ (x+1)

But (x^5 + 1)=(x+1)(x^4-x^3+x^2-x+1)

(x^5 + 1)/ (x+1)=(x+1)(x^4-x^3+x^2-x+1)/ (x+1)=

=(x^4-x^3+x^2-x+1)

**(x^8+x^6+x^4+x^2+1)/(x^4+x^3+x^2+x+1) = (x^4-x^3+x^2-x+1) q.e.d**