2019/002) Given $x^5+\frac{1}{x^5}$ Show that $x+\frac{1}{x}=3$ where x is real

let $x + \frac{1}{x} = y$ 
so cube both sides 
$x^3 + \frac{1}{x^3} + 3(x + \frac{1}{x}) = y^3$ 
or $x^3 + \frac{1}{x^3} = y^3-3y$ 
and $x^2+ \frac{1}{x^2} = y^2 -2$ 
multiply both to get 
$x^5 + \frac{1}{x} + x + \frac{1}{x^5} = 123 + y = (y^3-3y)(y^2-2) = y^5 - 5y^3 + 6y$ 
or $y^5 - 5y^3 + 5y - 123 =0$ 
or $(y-3)(u^4 + 3y^3 + 4y^2 + 12y + 41) = 0$ 
gives y = 3 or complex solutions 
so $x+ \frac{1}{x} = 3$ if x is real