a+b+c+d=0 
=> a + b = -(c+d) 
=> (a+b)^3 = -(c+d)^3 
=> a^3 + b^3 + 3ab (a+b) = -c^3- d^3 - 3cd (c+d) 
=> a^3+b^3+c^3+d^3 = -3ab (a+b) - 3cd (c+d) 
=> a^3 + b^3 + c^3 + d^3 = 3ab (c+d)- 3cd (c+d) [vì a+b = - (c+d)] 
==> a^3 + b^^3 + c^3 + d^3 =3 (c+d) (ab-cd) (đpcm)

Ta có :

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

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

\(\Rightarrow\left(b+c\right)^2=\left(a+d\right)^2\)

\(\Rightarrow\left(b+c\right)^2-\left(a+d\right)^2=0\)

\(\Rightarrow b^2+c^2+2bc-a^2-d^2-2ad=0\)

Lại có :

\(a^3+b^3+c^3+d^3\)

\(=\left(a+d\right)\left(a^2+d^2-ad\right)+\left(b+c\right)\left(b^2+c^2-bc\right)\)

\(=\left(b+c\right)\left(b^2+c^2-bc\right)-\left(b+c\right)\left(a^2+d^2-ad\right)\)

\(=\left(b+c\right)\left[\left(b^2+c^2-bc\right)-\left(a^2+d^2-ad\right)\right]\)

\(=\left(b+c\right)\left[\left(b^2+c^2+2bc-a^2-d^2-2ad\right)+3ad-3bc\right]\)

\(=\left(b+c\right)\left[0+3\left(ad-bc\right)\right]\)

\(=3\left(b+c\right)\left(ad-bc\right)\)

Vậy ...

  Ta có : a + b +c + d = 0

                  => a + d = - b - c

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

                => (a + d)³ = -(b + c)³

a³ + 3a²d + 3ad² + d³ = -(b³ + 3b²c + 3bc² + c³)

a³ + 3a²d + 3ad² + d³ = -b³ - 3b²c - 3bc² - c³

       a³ + b³ + c³ + d³ = -3a²d - 3ad² - 3b²c - 3bc² 

       a³ + b³ + c³ + d³ = -3ad(a + d) - 3bc(b + c)

       a³ + b³ + c³ + d³ = -3ad(-b - c) - 3bc(b + c)

       a³ + b³ + c³ + d³ = 3ad(b + c) - 3bc(b + c)

       a³ + b³ + c³ + d³ = 3(b + c)(ad - bc)

 a+b+c+d=0 
=>a+b = - (c+d) 
=> (a+b)^3= - (c+d)^3 
=> a^3 + b^3 + 3ab(a+b) = - c^3 - d^3 - 3cd(c+d) 
=> a^3 + b^3 + c^3 + d^3 = - 3ab(a+b) - 3cd(c+d) 
=> a^3 + b^3 + c^3 + d^3 = 3ab(c+d) - 3cd(c+d) ( Vì a+b = - (c+d)) 
==> a^3 + b^3 + c^3 + d^3 = 3(c+d)(ab-cd) (đpcm).

ta có : a+b+c+d=0 
=>a+b=-(c+d) 
=> (a+b)³=-(c+d)³ 
=> a³+b³+3ab(a+b)=-c³-d³-3cd(c+d) 
=> a³+b³+c³+d³=-3ab(a+b)-3cd(c+d) 
=> a³+b³+c³+d³=3ab(c+d)-3cd(c+d) ( vi a+b = - (c+d)) 
=> a³ +b³+c³+d³==3(c+d)(ab-cd)

(dpcm)

Ta có: a+b+c+d=0

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

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

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

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

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

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

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

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

 a+b+c+d=0 
=>a+b=-(c+d) 
=> (a+b)^3=-(c+d)^3 
=> a^3+b^3+3ab(a+b)=-c^3-d^3-3cd(c+d) 
=> a^3+b^3+c^3+d^3=-3ab(a+b)-3cd(c+d) 
=> a^3+b^3+c^3+d^3=3ab(c+d)-3cd(c+d) ( vi a+b = - (c+d)) 
==> a^3 +b^^3+c^3+d^3==3(c+d)(ab-cd) (dpcm)

Cho mk nói bạn Alan Walker chỉ là hs lớp 6 sao tài vậy

Nếu bạn ko biết làm thì thôi

Làm nhục anh em bạn ạ