Solving Polynomial Equations with Exponents of 40

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The discussion revolves around solving the polynomial equation x^4 - x^3 + x^2 - x + 1 = 0 and finding the value of x^40 - x^30 + x^20 - x^10 + x^0. Participants explore various methods, including symmetry, factoring, and the use of roots of unity, but find these approaches challenging. It is noted that the answer to the equation is 1, although the reasoning behind this is not straightforward. The conversation highlights the complexity of polynomial equations and the potential for alternative methods to simplify the problem.
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Polynomials help~~

Heh, so I posted this thread in the wrong category so I'm reposting it! =)
Hello. So here was this problem I came across:

If x^4-x^3+x^2-x^1+x^0=0, what is the numerical value of x^40-x^30+x^20-x^10+x^0?
I did try doing many stuffs (symmetry) & factoring, but I think none of these steps helped.
Enlighten the youngster, gracias.
 
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If 2X=0, what is 20X?
 


@adaptation -_- very funny, if it was a simple problem like that I wouldn't have posted it.
Btw, the answer's 1.
Hmm... but How...?
 


I found no shortcut in doing this :rolleyes: ,
maybe then the only way to do this is find the roots and put them in the second equation ,
Yea lengthy it is
 

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There could be a work around , using roots of unity , you need to think about it .
( sum of roots of unity and their properties )
 


our original polynomial is p(x) = x4-x3+x2-x+1.

note that p(-1) = 1 + 1 + 1 + 1 + 1 = 5, so -1 is not a root of p(x).

now consider q(x) = (x+1)p(x) = x5 + 1.

Mod note: rest of solution removed[/color]
 
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Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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