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 )

Dựa vào a^2 +b^2 = 1 và c^2+  d^2 = 1 và ac + bd +0

Ta có ab + cd = ab.1 + cd.1 = ab.(c^2 + d^2) + cd.(a^2+b^2)

                       = abc^2 + abd^2 + cda^2 + cdb^2

                       = ac(bc + da) + bd(ad + cb) = (ac+bd).(bc+da) = 0 . (bc+da) = 0

Vậy ab + cd =

\(b^2=ac\Rightarrow\frac{a}{b}=\frac{b}{c};c^2=bd\Rightarrow\frac{b}{c}=\frac{c}{d}\)

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

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

Ta lại có : \(\left(\frac{a}{b}\right)^3=\frac{a}{b}.\frac{a}{b}.\frac{a}{b}=\frac{a}{b}.\frac{b}{c}.\frac{c}{d}=\frac{a}{d}\) (2)

Từ (1) ; (2) => \(\frac{a}{d}=\left(\frac{a+b+c}{b+c+d}\right)^3\) (ĐPCM)