X is Real

Simple: You can't put x=0. now, If you put 'x' as a negative number, the left side of the equation becomes negative and If you put 'x' as a positive number, the left side of the equation becomes positive. So there is no solution of the equation in the real line.

Putting $a=(x+\frac{1}{x})$ We get, $a^3=-a$ $\Rightarrow a^2=-1, \space since \space a \ne 0$ for real $x$ .

Now this equation clearly can't be satisfied with any real number. Hence there are no real solutions

$With\quad x\quad \neq \quad 0$ $For\quad x\quad >\quad 0\quad we\quad have\quad :\quad { (x+\frac { 1 }{ x } ) }^{ 3 }+\quad (x+\frac { 1 }{ x } )\quad >\quad 0$ $For\quad x\quad <\quad 0\quad we\quad have\quad :\quad { (x+\frac { 1 }{ x } ) }^{ 3 }+\quad (x+\frac { 1 }{ x } )\quad <\quad 0$

So there is no reel roots for the equation

Substitute $\ y = x + \frac{1}{x}$ such that : $y^3 +y = y(y^2+1) = 0$ Clearly only y=0 has any chance of producing a real solution. However, it is clear to see that $x + \frac{1}{x} = \frac{x^2+1}{x} = 0$ which doesn't produce a real solution. Hence, there are no real solutions.

Here's an easy way to kill it. Make the substitution $u = x+\frac{1}{x}$ . Then we have $u^3=-u$ , or $u^2 =-1$ , which clearly admits no real solutions.

(x+(1/x))((x+(1/x))^2 +1)=0

By Zero Product Property, x + (1/x) = 0 and/or (x + (1/x))^2 = -1. Latter is not possible for real x.

Also x^2 + 1 = 0 is not possible for real x. Hence no real roots exist for this equation.

$(x+\frac{1}{x})^{3}+(x+\frac{1}{x})=0$ . As $x+\frac{1}{x} \ne 0$ , we have $(x+\frac{1}{x})^{2}=-1$ but an squared real number is never going to be negative.

If x+1/x=0 roots are imaginary Then, x^2+1/x^+3=0 Put t=x^2 and solve Roots are imaginary

(x+1/x)^3 +(x + 1/x)=0 , (x+1/x)((x +1/x)^2 + 1)=0 , (x^2 + 1)( (x + 1/x)^2 + 1) =0 , here in both bracket value can not be zero ,because these are positive .Due to this right hand side value can note be zero, this is grater than zero. therefore .there is no any real solution for given equation,

I solved it this way : Let f(x) = (x+ 1/x)^3 + (1 + 1/x) Expand and take LCM. Now find the number of sign changes of co-effecients , that tells the number of + roots , which is 0 here. Now find f(-x) , again repeat the same procedure. That is the number of -ve roots. which is 0 here . So all roots are complex , there are 6 complex roots.