If x² - 16x + 59 = 0, then what is the value of $${\left( {x - 6} \right)^2} + \frac{1}{{{{\left( {x - 6} \right)}^2}}}?$$

Solution (By Examveda Team)

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