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

Áp dụng t/c dtsbn:

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

Và \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\Rightarrow\dfrac{a^3}{b^3}=\dfrac{b^3}{c^3}=\dfrac{c^3}{d^3}=\dfrac{a^3+b^3+c^3}{b^3+c^3+d^3}\left(2\right)\)

\(\left(1\right),\left(2\right)\Rightarrow\dfrac{a^3+b^3+c^3}{b^3+c^3+d^3}=\left(\dfrac{a+b+c}{b+c+d}\right)^3\left(đpcm\right)\)

   Ta có   a + b + c + d = 0

\(\Leftrightarrow\)a+c = -( b+ d)

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

\(\Leftrightarrow\)a³ + c³ + 3ac.(a+c) = - [ b³ + d³ + 3bd( b+d) ]

\(\Leftrightarrow\)a³ + b³ + c³ + d³ = -3bd(b+d) - 3ac(a+c)

\(\Leftrightarrow\)a³ + b³ + c³ + d³ = -3bd( b+d) + 3ac( b+d)   

\(\Leftrightarrow\)a³ + b³ + c³ + d³ = 3( ac - bd)(b +d) (đpcm)

Ta có:     a + b + c +d = 0 => a + b + (c+d) = 0

=> a³ + b³ +(c+d)³ = 3ab(c+d)

=>a³ +b³ +c³ +d³ +3cd(c+d) = 3ab(c+d)

=> a³ +b³ +c³ +d³  = 3ab(c+d) – 3cd(c+d) = 3(c+d)(ab – cd).

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Giải:

\(a+b+c+d=0\)

\(\Leftrightarrow a+c=-b-d\)

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

Ta có:

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

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

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

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

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

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

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

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