If x² - 12x + 33 = 0, then what is the value of (x - 4)² + $$\frac{1}{{{{\left( {x - 4} \right)}^2}}}?$$

$$\eqalign{ & {x^2} - 12x + 33 = 0 \cr & {x^2} - 12x + 36 - 3 = 0 \cr & {\left( {x - 6} \right)^2} - 3 = 0 \cr & x = \sqrt 3 + 6 \cr & {\left( {x - 4} \right)^2} + \frac{1}{{{{\left( {x - 4} \right)}^2}}} \cr & \therefore {\left( {\sqrt 3 + 2} \right)^2} + \frac{1}{{{{\left( {\sqrt 3 + 2} \right)}^2}}} \cr & = {\left( {\sqrt 3 + 2} \right)^2} + {\left( {\sqrt 3 - 2} \right)^2} \cr & = 2\left( {{{\sqrt 3 }^2} + {2^2}} \right) \cr & = 2 \times 7 \cr & = 14 \cr} $$