Simplify the following expression:
$$\frac{{{{\left( {{a^2} - 4{b^2}} \right)}^3} + 64{{\left( {{b^2} - 4{c^2}} \right)}^3} + {{\left( {16{c^2} - {a^2}} \right)}^3}}}{{{{\left( {a - 2b} \right)}^3} + {{\left( {2b - 4c} \right)}^3} + {{\left( {4c - a} \right)}^3}}}$$

Solution(By Examveda Team)

$$\eqalign{ & \frac{{{{\left( {{a^2} - 4{b^2}} \right)}^3} + 64{{\left( {{b^2} - 4{c^2}} \right)}^3} + {{\left( {16{c^2} - {a^2}} \right)}^3}}}{{{{\left( {a - 2b} \right)}^3} + {{\left( {2b - 4c} \right)}^3} + {{\left( {4c - a} \right)}^3}}} \cr & {\text{Put }}a = b = c \cr & = \frac{{{{\left( { - 3{a^2}} \right)}^3} + 64{{\left( { - 3{a^2}} \right)}^3} + {{\left( {15{a^2}} \right)}^3}}}{{{{\left( { - a} \right)}^3} + {{\left( { - 2a} \right)}^3} + {{\left( {3a} \right)}^3}}} \cr & = \frac{{{a^6}\left[ { - 27 - 27 \times 64 + {{\left( {15} \right)}^3}} \right]}}{{{a^3}\left[ { - 1 - 8 + 27} \right]}} \cr & = \frac{{3{a^3}\left[ { - 9 - 576 + 1125} \right]}}{{18}} \cr & = \frac{{{a^3} \times 540}}{6} \cr & = 90{a^3} \cr & {\text{From option put }}a = b = c \cr & \left( {\text{A}} \right)\, - 45{a^3} \cr & \left( {\text{B}} \right)\,90{a^3} \cr & {\text{Hence option B is right answer}}{\text{.}} \cr} $$