Question

cho a+b+c = 0.  cmr :  a^3+b^3+c^3=3abc

Answer 1

Ta có:a+b+c=0
=>a+b=-c
=>(a+b)^3=(-c)^3
a^3+3a^2b+3ab^2+b^3=(-c)^3
=>a^3+3a^2b+3ab^2+b^3+c^3=0
=>a^3+b^3+c^3+3ab(a+b)=0
=>a^3+b^3+c^3+3ab(-c)=0
=>a^3+b^3+c^3-3abc=0
=>a^3+b^3+c^3=3abc (đpcm)

Answer 2

Ta có:

a^3+b^3+c^3−3abc=(a+b)^3−3a^2b−3ab^2+c^3−3abc=[(a+b)^3+c^3]−3ab(a+b+c)

=(a+b+c)[(a+b)^2−(a+b)c+c^2]−3ab(a+b+c)=0

-> a^3+b^3+c^3=3abc

Answer 3

a+b+c=0

=>(a+b+c)³=0

=>a³+b3+c³+3a²b+3ab²+3b²c+3bc²+3a²c+3ac²+6abc=0

=>a³+b³+c³+(3a²b+3ab²+3abc)+(3b²c+3bc²+3abc)+(3a²c+3ac²+3abc)-3abc=0

=>a³+b³+c³+3ab(a+b+c)+3ac(a+b+c)+3bc(a+b+c)=3abc

Do a+b+c=0

=>a³+b³+c³=3abc

Answer 4

hacker

Answer 5

  Ta có:  \(a+b+c=0 \)

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

\(\Leftrightarrow a^3+b^3+c^3+3a^2b+3ab^2+3b^2c+3bc^2+3c^2a+3ca^2+6abc=0\)

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

\(\Leftrightarrow a^3+b^3+c^3+3ab\left(a+b+c\right)+3bc\left(a+b+c\right)+3ca\left(a+b+c\right)=3abc\)

Mà \(a+b+c=0\) nên:

\(\Rightarrow a^3+b^3+c^3+0+0+0=3abc\)

\(\Leftrightarrow a^3+b^3+c^3=3abc\)(đpcm)