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\(\left\{{}\begin{matrix}b^2=ac\\c^2=bd\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{a}{b}=\dfrac{b}{c}\\\dfrac{b}{c}=\dfrac{c}{d}\end{matrix}\right.\Leftrightarrow\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\)

Đặt: \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=t\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a^3}{b^3}=\dfrac{b^3}{c^3}=\dfrac{c^3}{d^3}=\dfrac{a^3+b^3+c^3}{b^3+c^3+d^3}=t^3\\\dfrac{a}{b}.\dfrac{b}{c}.\dfrac{c}{d}=\dfrac{a}{d}=t^3\end{matrix}\right.\)

Ta có đpcm

Ta có :

\(b^2=ac\Leftrightarrow\dfrac{a}{b}=\dfrac{b}{c}\left(1\right)\)

\(c^2=bd\Leftrightarrow\dfrac{b}{c}=\dfrac{c}{d}\left(2\right)\)

Từ \(\left(1\right)+\left(2\right)\Leftrightarrow\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\)

\(\Leftrightarrow\dfrac{a^3}{b^3}=\dfrac{b^3}{c^3}=\dfrac{c^3}{d^3}\)

Áp dụng t,c dãy tỉ số bằng nhau ta có :

\(\dfrac{a^3}{b^3}=\dfrac{b^3}{c^3}=\dfrac{c^3}{d^3}=\dfrac{a^3+b^3+c^3}{b^3+c^3+d^3}\left(3\right)\)

Lại có :

\(\dfrac{a^3}{b^3}=\dfrac{b^3}{c^3}=\dfrac{c^3}{d^3}=\dfrac{a}{b}.\dfrac{b}{c}.\dfrac{c}{d}=\dfrac{a}{d}\left(4\right)\)

Từ \(\left(3\right)+\left(4\right)\Leftrightarrowđpcm\)

cái 

d2 = ac , c2 =bd sai hay đúng đó

b² = ac

\(\Rightarrow\frac{a}{b}=\frac{b}{c}\)( 1 )

c² = bd

\(\Rightarrow\frac{b}{c}=\frac{c}{d}\)( 2 )

Từ ( 1 ) và ( 2 ) \(\Rightarrow\frac{a}{b}=\frac{b}{c}=\frac{c}{d}\)

\(\Rightarrow\frac{a^3}{b^3}=\frac{b^3}{c^3}=\frac{c^3}{d^3}=\frac{a^3+b^3+c^3}{b^3+c^3+d^3}=\frac{a}{b}.\frac{b}{c}.\frac{c}{d}=\frac{a}{d}\)

\(\hept{\begin{cases}b^2=ac\\c^2=bd\end{cases}}\Rightarrow\hept{\begin{cases}\frac{a}{b}=\frac{b}{c}\\\frac{b}{c}=\frac{c}{d}\end{cases}\Rightarrow\frac{a}{b}=\frac{b}{c}=\frac{c}{d}\Rightarrow\frac{a^3}{b^3}=\frac{b^3}{c^3}=\frac{c^3}{d^3}}\)

Áp dụng tính chất của dãy tỉ số bằng nhau ta có:

\(\frac{a^3}{b^3}=\frac{b^3}{c^3}=\frac{c^3}{d^3}=\frac{a^3+b^3+c^3}{b^3+c^3+d^3}\left(1\right)\)

Lại có: \(\frac{a^3}{b^3}=\frac{a}{b}\cdot\frac{a}{b}\cdot\frac{a}{b}=\frac{a}{b}\cdot\frac{b}{c}\cdot\frac{c}{d}=\frac{a}{d}\left(2\right)\)

Từ (1) và (2) => đpcm

b²=ac=>\(\frac{a}{c}=\frac{b}{c}\left(1\right)\)

c²=bd=>\(\frac{b}{c}=\frac{c}{d}\left(2\right)\)

từ (1) và (2) => \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}\Rightarrow\frac{a^3}{b^3}=\frac{c^3}{d^3}=\frac{b^3}{c^3}\Rightarrow\frac{a^3}{b^3}=\frac{a}{b}=\frac{b}{c}=\frac{c}{d}^{^{ }}\left(\text{(*)}\right)\)

\(\frac{a^3}{b^3}=\frac{b^3}{c^3}=\frac{c^3}{d^3}\)= \(\frac{a^3+b^3+c^3}{b^3+c^3+d^3}\)           (**)

Từ (*) và (**) => \(\frac{a.b.c}{b.c.d}=\frac{a}{d}\)

Ta có:\(a+b+c+d=0\)

\(a+c=-\left(b+d\right)\)

\(\left(a+c\right)^3=-\left(b+d\right)^3\)

\(\Leftrightarrow a^3+c^3+3ac\left(a+c\right)=-\left[b^3+d^3+3bd\left(b+d\right)\right]\)

\(\Leftrightarrow a^3+b^3+c^3+d^3=-3bd\left(b+d\right)-3ac\left(a+c\right)\)

\(\Leftrightarrow a^3+b^3+c^3+d^3=-3bd\left(b+d\right)+3ac\left(b+d\right)\)

\(\Leftrightarrow a^3+b^3+c^3+d^3=3\left(ac-bd\right)\left(b+d\right)\left(đpcm\right)\)

Sửa đề một chút : Cmr a³ + b³ + c³ + d³ = 3 ( ac - bd ) ( b + d ) 

a + b + c + d = 0 

=> a + c = - ( b + d )

\(\Leftrightarrow\left(a+c\right)^3=-\left(b+d\right)^3\)

\(\Leftrightarrow a^3+3a^2c+3ac^2+c^3=-b^3-d^3-3b^2d-3bd^2\)

\(\Leftrightarrow a^3+3ac\left(a+c\right)+c^3=-b^3-d^3-3bd\left(b+d\right)\)

\(\Leftrightarrow a^3+b^3+c^3+d^3=-3ac\left(a+c\right)-3bd\left(b+d\right)\)

\(\Leftrightarrow a^3+b^3+c^3+d^3=3ac\left(b+d\right)-3bd\left(b+d\right)\)

\(\Leftrightarrow a^3+b^3+c^3+d^3=3\left(ac-bd\right)\left(b+d\right)\)( đpcm )