If $\frac{1}{(x-2)} + x = 8 $, then what is the value of $\frac{1}{(x-2)^2}+(x - 2)^2 $ ?

38 36 40 34

34

If $K+\frac{1}{K}=n$ then, $K^2+\frac{1}{K^2}$ = n2 – 2 If $\frac{1}{(x-2)} + x = 8 $, then what is the value of $\frac{1}{(x-2)^2}+(x - 2)^2 $ ? Subtract 2 from both the sides, $\frac{1}{(x-2)} + (x-2) = 8-2 $, $\frac{1}{(x-2)} + (x-2) = 6 $, $\frac{1}{(x-2)^2}+(x - 2)^2 $ = 62 – 2 = 34