# Verify if (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)

sciencesolve | Teacher | (Level 3) Educator Emeritus

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You also may use distributive property of multiplication over addition, such that:

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

`x^4*(x^4+x^3+x^2+x+1) - x^3*(x^4+x^3+x^2+x+1)`

`+ x^2*(x^4+x^3+x^2+x+1) - x*(x^4+x^3+x^2+x+1)`

`+ 1*(x^4+x^3+x^2+x+1)`

`(x^4-x^3+x^2-x+1)*(x^4+x^3+x^2+x+1)`

`= x^8 + x^7 + x^6 + x^5 + x^4 - x^7`

`- x^6 - x^5 - x^4 - x^3 + x^6 + x^5 + x^4`

`+ x^3 + x^2 - x^5 - x^4 - x^3 - x^2 - x`

`+ x^4 + x^3 + x^2 + x + 1`

Reducing duplicate members yields:

`(x^4-x^3+x^2-x+1)*(x^4+x^3+x^2+x+1)`

`= x^8 + x^6 + x^4 + x^2 + 1`

Hence, testing if the given expression holds, performing the multiplcations and reducing the duplicate members, yields that `(x^4-x^3+x^2-x+1)*(x^4+x^3+x^2+x+1) = x^8 + x^6 + x^4 + x^2 + 1` holds.

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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We’ll use the identity:

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

We'll divide by (x^4+x^3+x^2+x+1) both sides:

(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’ve noticed that both, denominator and numerator of the left side ratio, represent the second factor from expanding of a^n – b^n.

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) is verified.