Here goes the question
Assuming "x" is a real number,such that the equation is given $(x+\frac {1} {x})^2=3$,to evaluate $x^3+\frac {1} {x^3}$
And here goes my working,taking 2+ hours,only to give up solving 'x'
$x^2+2+\frac {1} {x^2}=3$ [$Note:(a+b)^2= a^2+2ab+b^2$]
$x^2+\frac {1} {x^2}=3-2$
$x^2+\frac {1} {x^2}=1$ (Compares to $x^3+\frac {1} {x^3}$)
$(x^2+\frac {1} {x^2})(x+\frac {1} {x})=1(x+\frac {1} {x})$
$x^3+\frac {1} {x^3}=x+\frac {1} {x}-x-\frac {1} {x}$
So,$x^3+\frac {1} {x^3}$ is literally equals to "0".Im not convinced,as I find it weird the answer to be '0'.
So my question is,other than connecting $(x+\frac {1} {x})^2$ to $x^3+\frac {1} {x^3}$,from $(x+\frac {1} {x})^2=3$,how do you solve 'x'?(Because finding "x" it's easier to check,especially by substitution)
You are right to think something fishy is going on. In fact there is no real number $x$ such that $(x+\frac{1}{x})^2=3$.
(Incidentally, there are four complex solutions for $x$, and all of them do have the property $x^3 + \frac{1}{x^3} = 0$, as you proved.)