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If \(\Large{x+\frac{1}{x}=2}\),then \(\Large{x^2+\frac{1}{x^8}=?}\)
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@mathslover @mukushla @sauravshakya plz help :)
solve the first equation for x
isn't it x^8 +1/x^8 ?
ok!\[x=2-1/x\]
\[x^2-2x+1=0\]
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