If $$x + \frac{1}{{16x}} = 3,$$   then the value of $$16{x^3} + \frac{1}{{256{x^3}}}$$   is:

Solution (By Examveda Team)

$$\eqalign{ & x + \frac{1}{{16x}} = 3 \cr & 2x + \frac{1}{{8x}} = 6 \cr & {\text{Cube both side}} \cr & 8{x^3} + \frac{1}{{512{x^3}}} + 3 \times 2 \times \frac{1}{8} \times 6 = 216 \cr & 8{x^3} + \frac{1}{{512{x^3}}} = 216 - \frac{9}{2} \cr & {\text{Multiply by '2' both side}} \cr & 16{x^3} + \frac{1}{{256{x^3}}} = 432 - 9 = 423 \cr} $$