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?

Logic Oct 10, 2019

#1**+2 **

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

.Rom Oct 10, 2019

#3**+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