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:

Correct option is C

Given:

​$$\frac{a}{b} = \frac{b}{c} = \frac{c}{d} = k$$​

This implies:

​$$a = bk, \quad b = ck, \quad c = dk$$​

We need to evaluate:

$$\frac{b^3 + c^3 + d^3}{a^3 + b^3 + c^3}$$

Solution:

Substituting a, b, c in terms of d:

​$$a = k^3 d, \quad b = k^2 d, \quad c = k d$$​

Substituting these values in the given fraction:  

$$\frac{b^3 + c^3 + d^3}{a^3 + b^3 + c^3}$$ 

= $$\frac{ (k^2d)^3 + (kd)^3 + d^3}{(k^3d)^3 + (k^2d)^3 + (kd)^3}$$   

=  $$\frac{d^3(k^6 + k^3 + 1)}{d^3k^3( k^6 + k^3 + 1)}$$-​

​​$$=\frac{k^6 + k^3 + 1}{k^3 (k^6 + k^3 + 1)} =\frac{1}{k^3} =\frac{d}{a}$$

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