If $$\frac{a}{b} = \frac{b}{c} = \frac{c}{d},$$   then $$\frac{{{b^3} + {c^3} + {d^3}}}{{{a^3} + {b^3} + {c^3}}}$$   will be equal to -

$$\eqalign{ & {\text{Let }}\frac{a}{b} = \frac{b}{c} = \frac{c}{d} = k \cr & {\text{Then,}} \cr & a = bk, \cr & b = ck, \cr & c = dk \cr & {\text{Also,}} \cr & \Rightarrow \frac{a}{b} \times \frac{b}{c} \times \frac{c}{d} = {k^3} \cr & \Rightarrow {k^3} = \frac{a}{{d}} \cr & \therefore \frac{{{b^3} + {c^3} + {d^3}}}{{{a^3} + {b^3} + {c^3}}} \cr & = \frac{{{b^3} + {c^3} + {d^3}}}{{{{\left( {bk} \right)}^3} + {{\left( {ck} \right)}^3} + {{\left( {dk} \right)}^3}}} \cr & = \frac{{{b^3} + {c^3} + {d^3}}}{{{k^3}\left( {{b^3} + {c^3} + {d^3}} \right)}} \cr & = \frac{1}{{{k^3}}} \cr & = \frac{d}{a} \cr} $$