If 4x² - 6x + 1 = 0, then the value of $$8{x^3} + \frac{1}{{8{x^3}}}$$  is:

Solution (By Examveda Team)

$$\eqalign{ & 4{x^2} - 6x + 1 = 0 \cr & 2x\left( {2x - 3 + \frac{1}{{2x}}} \right) = 0 \cr & 2x + \frac{1}{{2x}} = 3 \cr & 8{x^3} + \frac{1}{{8{x^3}}} = 27 - 3 \times 3 \cr & 8{x^3} + \frac{1}{{8{x^3}}} = 18 \cr} $$