The 3rd term from the end in the expansion of $( x + \frac {1}{x} )^6$is

$\textbf{Step 1: Find position of term form beginning}$

$\text{Given expansion is}$ $( x + \dfrac {1}{x} )^6$

$\text{If n= power of the expansion and p is position of term from end then,}$

$\text{pth term from the end = ( n- p +2)th term from the beginning}$.

$\therefore \text{3rd term from the end= (6 -3 +2)th term }$

$\text{= 5th term from beginning}$

$\textbf{Step 2: Find value of 5 th term from expansion}$

$\text{General term of expansion$(p+y)^n$ is given by-}$

$T_n= T_{r +1 }=^nC_r p^{ (n-r)}.y^r$

$\text{Comparing given equation with standard form, p=x and y=1/x}$

$\Rightarrow T_5 = T_{ 4+1} = ^6C_4. x^{(6-4)} . ( \dfrac {1}{x})^4.$

$=^6C_4 .x^2 .\dfrac {1}{x^4}$

$= \dfrac { 6 \times 5}{2 \times 1} \times \dfrac {1}{x^2}$

$= \dfrac {15}{x^2}$

$\textbf{Therefore, 3rd term from end in expansion is }$

$\mathbf{= \dfrac {15}{x^2} }$