Ta có:

$b^2 = ac$

$\to \dfrac ba =\dfrac cb$

$c^2 = bd$

$\to \dfrac cb =\dfrac dc$

$\to \dfrac ba =\dfrac cb =\dfrac dc$

Đặt $\dfrac ba =\dfrac cb =\dfrac dc = k$

$\to \begin{cases}b = ka\\c = kb\\d = kc\end{cases}$

$\to \dfrac ba \cdot\dfrac cb \cdot\dfrac dc = k^3$

$\to \dfrac da = k^3\qquad (1)$

Mặt khác:

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

$=\dfrac{k^3a^3 + k^3b^3 + k^3c^3}{a^3 + b^3 + c^3}$

$=\dfrac{k^3(a^3 + b^3 + c^3)}{a^3 + b^3 + c^3}$

$= k^3\qquad(2)$

$(1)(2)\to \dfrac{b^3 + c^3 + d^3}{a^3 + b^3 + c^3} = \dfrac da$

$b^{2}$=ac

=> bb=ac =>$\frac{b}{c}$ = $\frac{a}{b}$ =>$$\frac{b}{c}$^{3}$ =$ $\frac{a}{b}$ ^{3}$    (1)

=> cc=bd =>$\frac{c}{d}$ = $\frac{b}{c}$  =>$$\frac{c}{d}$^{3}$ = $$\frac{b}{c}$^{3}$    (2)

Từ (1) và (2)

=> $\frac{b}{c}$ = $\frac{a}{b}$=$\frac{c}{d}$

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

Từ $$\frac{b}{c}$^{3}$

=>$$\frac{b}{c}$^{3}$=$\frac{a}{b}$ .$\frac{a}{b}$.$\frac{a}{b}$

                                   =