If x^2+x+1=0, then the value of (x+(1)/(x))^4+(x^2+1x^2)^4+(x^3+1x^3)^4+ +(x^25+1x^25)^4 i…

Question

If ${x}^{2}+x+1=0$, then the value of ${(x+\frac{1}{x})}^{4}+{({x}^{2}+\frac{1}{{x}^{2}})}^{4}+{({x}^{3}+\frac{1}{{x}^{3}})}^{4}+\ldots +{({x}^{25}+\frac{1}{{x}^{25}})}^{4}$ is:

Accepted Answer

$\begin{gathered} {(w+\frac{1}{w})}^{4}+{({w}^{2}+\frac{1}{{w}^{2}})}^{4}+{({w}^{3}+\frac{1}{{w}^{3}})}^{4}+\ldots \ {({w}^{25}+\frac{1}{{w}^{25}})}^{4} \end{gathered}$

$\begin{gathered} \sum {({w}^{k}+\frac{1}{{w}^{k}})}^{4} \ K=3x\Rightarrow {w}^{3x}+\frac{1}{{w}^{3x}}=2 \ k
e 3x\Rightarrow {w}^{k}+\frac{1}{{w}^{k}}=-1 \ \sum _{k=1}^{25}{({w}^{k}+\frac{1}{{w}^{k}})}^{4}\Rightarrow 8(1+1+{2}^{4})+1 \ =145 \end{gathered}$

Solutions · 1