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If $x+\dfrac{1}{x}=1$ then evaluate $x^{29}+\dfrac{1}{x^{89}}$
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## 1 Answer

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Answer:

$1$

Explanation:

$x+\frac{1}{x}=1$

$\Rightarrow \quad x^{2}+1=x$

$\Rightarrow x^{2}-x+1=0$

$\Rightarrow(x-1)\left(x^{2}-x+1\right)=0$

$\Rightarrow x^{3}-1=0$

$\Rightarrow x^{3}=1 .$

Given the expression:
$x^{29}+\dfrac{1}{x^{89}}$

$=\dfrac{x^{30}}{x}+\dfrac{x}{x^{90}}$

$=\dfrac{\left(x^{3}\right)^{10}}{x}+\dfrac{x}{\left(x^{3}\right)^{30}}$

$=\dfrac{1}{x}+\frac{x}{1}$

$=\dfrac{1}{x}+x=1$

by Diamond (38,951 points)
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