x^6+x^5-2x^4-x^3+x^2+1=0

Y=x^4-x^2+x-1

show y satisfies a quadratic equation with rational coefficients.

I've tried and have not be able to do so, so far.

Do it, please.

Thanks.

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- Aug 29th 2010, 09:47 AM #1

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- Aug 2010
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- Aug 29th 2010, 10:53 AM #2
I think this is just an exercise in division of polynomials. Square y, then divide by the expression that is given to be zero:

. . . . .$\displaystyle \begin{aligned}y^2 &= (x^4-x^2+x-1)^2 \\

&= x^8 - 2x^6 + 2x^5 - x^4 - 2x^3 + 3x^2 - 2x + 1 \\

&= (x^2 - x + 1)(x^6 + x^5-2x^4-x^3+x^2+1) - x^4 + x^2 - x \\

&= -y - 1\end{aligned}$

Therefore y satisfies the quadratic equation $\displaystyle y^2+y+1=0$.

(I'll leave you to convince yourself that this is an occasion where division by zero is justified!)