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avatar+1195 

Given that x is real and \(x^3+\frac{1}{x^3}=52\), find \(x+\frac{1}{x}\).

 

I got 6 but according to AoPs it isn't right?

 Oct 10, 2019
 #1
avatar+6046 
+2

\(\left(x+\dfrac 1 x\right)^3 = \\ x^3 + 3x+ 3x^{-1}+x^{-3} = \\ 52 + 3\left(x + \dfrac 1 x\right)\\ u=\left(x+\dfrac 1 x\right)\\ u^3 - 3u - 52=0\\ (u-4) \left(u^2+4 u+13\right) = 0\\ \text{The left hand factor is zero at $x=4$, the right hand factor has complex roots}\\ \text{The solution you are after is $x=4$}\)

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 Oct 10, 2019
 #3
avatar+2505 
+2

Holy cow rom that is smart

 

I always thought that your profile pic was a cow until I looked closely and it was a horse, er, or maybe a pony.

CalculatorUser  Oct 10, 2019
edited by CalculatorUser  Oct 10, 2019
 #5
avatar+6046 
+2

It's a miniature pony.  It's not mine.  Just a rando from youtube.

 

Apparently this site doesn't like us posting youtube links as my previous reply went into moderation limbo.

Rom  Oct 10, 2019
 #2
avatar+1195 
+2

thanks rom!

 Oct 10, 2019

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