\(a+b+c+d=0\Rightarrow a+b=-\left(c+d\right)\)

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

\(\Rightarrow\left(a+b\right)^3+\left(c+d\right)^3=0\)

\(\Rightarrow a^3+b^3+3ab\left(a+b\right)+c^3+d^3+3cd\left(c+d\right)=0\)

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

\(\Rightarrow a^3+b^3+c^3+d^3=3ab\left(c+d\right)-3cd\left(c+d\right)\) (do \(a+b=-\left(c+d\right)\)

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

Theo đề, a+b+c+d=0

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

Ta có: \(VT=\left(a+b\right)\left(a^2-ab+b^2\right)+\left(c+d\right)\left(c^2-cd+d^2\right)\)

\(\Leftrightarrow VT=\left(c+d)\left(c^2-cd+d^2-a^2+ab-b^2\right)\right)\)

Để có ĐPCM ta xét hiệu: \(c^2-cd+d^2-a^2+ab-b^2-3\left(ab+cd\right)=c^2-4cd+d^2-a^2-2ab-b^2=c^2-4cd+d^2-\left(a+b\right)^2=c^2-4cd+d^2-\left(c+d\right)^2=-6cd\)

S nó ko = 0 ta:::xem lại đề..Hay mk lm sai j đó

\(a.a^3+b^3+c^3=3abc\)

⇔ \(a^3+b^3+c^3-3abc=0\)

⇔ \(\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=0\)

⇔ \(\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)=0\)

⇔\(\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)=0\)

⇔ \(\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)

Với : a + b + c = 0 thì dễ thấy đẳng thức trên đúng .

Từ đó suy ra : đpcm .

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

⇔ \(a+b=-\left(c+d\right)\)

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

⇔ \(a^3+b^3+3a^2b+3ab^2=-\left(c^3+3c^2d+3cd^2+d^3\right)\)

⇔ \(a^3+b^3+c^3+d^3=-3c^2d-3cd^2-3a^2b-3ab^2\)

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

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

⇔ \(a^3+b^3+c^3+d^3=3\left(c+d\right)\left(ab-cd\right)\) ( đpcm)

1) Áp dụng BĐT bun-hi-a-cốp-xki ta có:

\(\left(a+d\right)\left(b+c\right)\ge\left(\sqrt{ab}+\sqrt{cd}\right)^2\)

\(\Leftrightarrow\sqrt{\left(a+d\right)\left(b+c\right)}\ge\sqrt{ab}+\sqrt{cd}\)( vì a,b,c,d dương )

Dấu " = " xảy ra \(\Leftrightarrow\frac{a}{b}=\frac{c}{d}\)

ab=bc=cd

nên a=b=c=d

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

Ta có: a+b+c+d=0

⇔\(a+d=-\left(b+c\right)\)

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

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

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

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

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

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

\(\Leftrightarrow a^3+b^3+c^3+d^3=\left(a+d\right)\cdot3\cdot\left(-ad+bc\right)\)

\(\Leftrightarrow a^3+b^3+c^3+d^3=-\left(b+c\right)\cdot3\cdot\left[-\left(ad-bc\right)\right]\)

\(\Leftrightarrow a^3+b^3+c^3+d^3=3\cdot\left(b+c\right)\cdot\left(ad-bc\right)\)(đpcm)