[tex]\left( x+\frac{1}{x}\right)^{8} =\sum _{k=0}^{8}\binom{8}{k} \cdotp x^{8-k} \cdotp \left(\frac{1}{x}\right)^{k}\\\\\\ \begin{array}{l}\text{We\ wish\ to\ find\ } x^{0} :\\\\x^{8-k} \cdotp \left(\frac{1}{x}\right)^{k} =x^{0}\\\\\Longrightarrow x^{8-k} \cdotp x^{-k} =x^{0}\\\\\Longrightarrow \cancel{x}^{8-2k} =\cancel{x}^{0}\\\\\Longrightarrow 2k=8\\\\\Longrightarrow k=4\end{array}\\\\\therefore \binom{8}{4} \cdotp x^{4} \cdotp \left(\frac{1}{x}\right)^{4} =\frac{8!}{( 8-4) !\cdotp 4!}[/tex]
[tex]\binom{8}{4} \cdotp x^{4} \cdotp \left(\frac{1}{x}\right)^{4} =\frac{8\cdotp 7\cdotp 6\cdotp 5\cdotp \cancel{4!}}{\cancel{4!} \cdotp 4\cdotp 3\cdotp 2\cdotp 1}\\\\\binom{8}{4} \cdotp x^{4} \cdotp \left(\frac{1}{x}\right)^{4}=70\\\\\therefore \boxed{\boxed{\text{The\ constant\ term\ is\ } 70}}[/tex]
