Question

cho a+b+c+d=0.   Cm:

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

Answer 1

ko biết

Answer 2

ta có

a+b+c+d=0

=> b+c=-(a+d) => (b+c)³=-(a+d)³

=> b³+c³+3bc(b+c)= -[a³+d³+3ad(a+d)]

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

=3(b+c)(ad-bc)

Answer 3

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

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

\(\Rightarrow\text{(b+c)^3=-(a+d)^3}\)

\(\text{\Rightarrow b^3+c^3+3bc(b+c)}\)

\(\text{= -[a^3+d^3+3ad(a+d)]}\)

\(\text{\Rightarrow a^3+b^3+c^3+d^3=-3ad(a+d)-3bc(b+c)= 3ad(b+c)-3bc(b+c)}\)

\(\text{=3(b+c)(ad-bc)}\)