Question

Chứng minh rằng : a+b+c+d=0 thì a³+b³+c³+d³=3(ac-bd)(b+d)

Answer 1

   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)

 

 

 

Answer 2

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).