If x2 - 16x + 59 = 0, then what is the value of ( x - 6 )^2 + 1 ( x - 6 )^2?

Answer: Option B

$$\begin{array}{rl}& x^2-16x+59=0\\ & {\left(x-6\right)}^2+\frac{1}{{\left(x-6\right)}^2}=?\\ & \text{Let }x-6=t\\ & x=t+6\\ & t^2+\frac{1}{t^2}=?\\ & {\left(t+6\right)}^2-16\left(t+6\right)+59=0\\ & t^2+12t-36-16t-96+59=0\\ & t^2-4t-1=0\\ & \text{Dividing by }t,\text{ we get}\\ & t-4-\frac{1}{t}=0\\ & t-\frac{1}{t}=4\\ & \text{By squaring, we get}\\ & t^2+\frac{1}{t^2}-2=16\\ & t^2+\frac{1}{t^2}=18\end{array}$$