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Q: If $x^2+\frac{1}{x^2}=7$, find the value of $\frac{x^6+1}{x^3}$.

Given,
$x^2+\frac{1}{x^2}=7$
$\rightarrow$ $(x+\frac{1}{x})^2=9$
$\rightarrow$ $(x+\frac{1}{x})=\pm 3$

Is it permissible to ignore $-3$ and use only $+3$ to find the value of $\frac{x^6+1}{x^3}$ as it is done in this website - https://www.toppr.com/ask/question/if-x2-dfrac1x2-7-find-the-value-of-x3dfrac1x3/

Russell
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    Hello :) No, it's not ok to ignore that. Maybe you have more than one solution. Observe $x^{n+1}+\frac{1}{x^{n+1}}=(x^n+\frac{1}{x^n}) (x+\frac{1}{x}) -(x^{n-1}+\frac{1}{x^{n-1}}) $ – Jochen Mar 11 '22 at 17:55
  • Hi @Jochen, thanks for your comment. Sorry, I don't understand it, it's a little advance for me. Could you please break it down a little? – Russell Mar 11 '22 at 18:00
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    Sure :) You are allowed to ignore $-3$, if $x>0$ is given. Otherwise, there are at least two solutions. Let $s_n:=x^n+\frac{1}{x^n}$ for all $n\in \mathbb N$. Then $s_0=2$, $s_1=\pm 3$ and $s_{n+2}=s_{n+1}s_1-s_n$. – Jochen Mar 11 '22 at 18:07
  • Thank you so much for the explanation. – Russell Mar 11 '22 at 18:09
  • You're welcome :) – Jochen Mar 11 '22 at 18:10
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    @Jochen Well, in general, if you have two solutions, this must be acknowledged. But for this particular problem, it can be shown that it only matters what the absolute of $s_1.$ For instance, you mentioned the equation $$s_{n+2}-s_1s_{n+1}+s_n=0.$$ Let $s_n=t^n,$ hence $t^2-s_1t+1=0.$ This implies $$t=\frac{s_1}2\pm\sqrt{\left(\frac{s_1}2\right)^2-1}.$$ For the undetermined coefficients, notice that $2=A+B$ and $s_1=A\left[\frac{s_1}2-\sqrt{\left(\frac{s_1}2\right)^2-1}\right]+B\left[\frac{s_1}2+\sqrt{\left(\frac{s_1}2\right)^2-1}\right].$ – Angel Mar 11 '22 at 18:50
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    The latter equation implies $B=A=1.$ So $$s_n=\left[\frac{s_1}2+\sqrt{\left(\frac{s_1}2\right)^2-1}\right]^n+\left[\frac{s_1}2-\sqrt{\left(\frac{s_1}2\right)^2-1}\right]^n.$$ Let $r_n$ correspond to the sequence where $s_1$ is replaced with $-s_1.$ Then $r_n=\left[-\frac{s_1}2+\sqrt{\left(\frac{s_1}2\right)^2-1}\right]^n+\left[-\frac{s_1}2-\sqrt{\left(\frac{s_1}2\right)^2-1}\right]^n=(-1)^ns_n.$ – Angel Mar 11 '22 at 18:55

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