Question 1035971
a, b, c, d in continued proportion ==> a/b = b/c = c/d = k for some constant of proportionality k 

==> {{{a = bk = ck^2 = dk^3}}}

==> {{{b = a(1/k)}}}, {{{c = a(1/k)^2}}}, and {{{d = a(1/k)^3}}}, and so a,b,c, and d form a geometric sequence.

==>{{{(a^3+b^3+c^3)/(b^3+c^3+d^3) = (a^3+a^3*(1/k^3)+a^3*(1/k^6))/(a^3*(1/k^3)+a^3*(1/k^6)+a^3*(1/k^9)) = (1+1/k^3+1/k^6)/(1/k^3+1/k^6+1/k^9) =1/( 1/k^3) = k^3}}}.

BUT, {{{a = dk^3}}}, and so {{{k^3 = a/d}}}, and the statement is proved.