If $$2x + \frac{2}{{9x}} = 4{\text{,}}$$   then the value of $$27{x^3} + \frac{1}{{27{x^3}}}$$   is?

$$\eqalign{ & 2x + \frac{2}{{9x}} = 4 \cr & {\text{Multiply by }}\frac{3}{2}{\text{ on both sides}} \cr & \Rightarrow 3x + \frac{1}{{3x}} = 6 \cr & {\text{Taking cube on both sides}} \cr & \Rightarrow {\left( {3x + \frac{1}{{3x}}} \right)^3} = {6^3} \cr & \Rightarrow 27{x^3} + \frac{1}{{27{x^3}}} + 3 \times 3x \times \frac{1}{{3x}}\left( {3x + \frac{1}{{3x}}} \right) = 216 \cr & \Rightarrow 27{x^3} + \frac{1}{{27{x^3}}} + 3 \times 6 = 216 \cr & \Rightarrow 27{x^3} + \frac{1}{{27{x^3}}} = 216 - 18 \cr & \Rightarrow 27{x^3} + \frac{1}{{27{x^3}}} = 198 \cr} $$