$x^g$+$\frac{1}{x^a}=x^g+x^{-a}$
Since g = a,
$x^g+\frac{1}{x^a}=x^a+x^{-a}$
For the expression  $x^a+x^{-a}$ to be a polynomial, a and -a must be whole number so that the expression $\frac{1}{x^a}$ = 1 , hence a should be equal to 0 .
Thus, a = 0.
When a=0, the given polynomial will get simplified further as:
$x^g+\frac{1}{x^a}=x^a+x^{-a}$
$\Rightarrow$ $x^g+\frac{1}{x^a}=x^0+x^{-0}$
$\Rightarrow$ $x^g+\frac{1}{x^a}=x^0+x^0$
$\Rightarrow$ $x^g+\frac{1}{x^a}=1+1=2$
'2' is a constant polynomial in one variable with degree zero . Because $2=2x^0$.
So, 'a' must be equal to zero so that given expression is a polynomial in variable.