Show that $x^a+x-b=0$ must have only one positive real root and not exceed the $\sqrt[a]{b-1}$ If we take the equation $$x^3+x-3=0$$ and solve it to find the real roots, we will get only one positive real roots which is $(x=1.213411662)$. If we comparison this with $\sqrt[3]{3-1}=1.259921$, we will find that $x$ is less than $\sqrt[3]{3-1}$.This always happens with any value of $a$ so that $a$ any positive real number  and $b$ is a positive real number.
So we can ask the following:
1-prove that $x^a+x-b=0$ must have only one positive real root if $a$ positive real number  and $b$ is a positive real number greater than $1$. 
2-The value of this roots must be less than $\sqrt[a]{b-1}$ 
3-What happens  when $a$  is a complex value.I mean this thing stays right or not?

If $a > 0$, you can use the Intermediate Value Theorem.
If $a < 0$, let $v$ be the minimum value of $x^a + x$ on $(0,\infty)$.  Then $x^a + x - b$ has no positive real roots if $b < v$, one if $b = v$ and two if $b > v$.
If $a$ is complex, $x^a + x - b$ is  real only for a discrete set of positive real $x$'s, so there will usually be no positive real roots.
