a +b +c = 0 => a + b = -c
ta có
a^3 + b^3 + c^3 = ( a + b )^3 - 3ab(a+b) + c^3
= -c^3 - 3ab.-c + c^3
= 3abc
=> ĐPCM
a^3 + b^3 = ( a+b)^3 - 3ab(a+b)
a^3 + b^3 + c^3 = ( a+b )^3 - 3ab(a+b) + c^3
a^3 +b^3 + c^3 = (-c)^3 - 3ab(-c) +c^3
a^3 + b^3 +c^3 = 3abc
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)+3bc(a+b+c)+3ac(a+b+c)=3abc
Do a+b+c=0
=>a3+b3+c3=3abc(ĐPCM)
\(a+b+c=0\Rightarrow a+b=-c\)
\(\Leftrightarrow\left(a+b\right)^3=\left(-c\right)^3\)
\(\Leftrightarrow a^3+3a^2b+3ab^2+b^3=-c^3\)
\(\Leftrightarrow a^3+b^3+c^3=-3a^2b-3ab^2\)
\(\Leftrightarrow a^3+b^3+c^3=-3ab\left(a+b\right)\)
\(\Leftrightarrow a^3+b^3+c^3=-3ab.\left(-c\right)\)
\(\Leftrightarrow a^3+b^3+c^3=3abc\left(đpcm\right)\)
Bài làm :
\(a+b+c=0\Leftrightarrow a+b=-c\)
\(\Leftrightarrow\left(a+b\right)^3=\left(-c\right)^3\)
\(\Leftrightarrow a^3+3a^2b+3ab^2+b^3=-c^3\)
\(\Leftrightarrow a^3+b^3+c^3=-3a^2b-3ab^2\)
\(\Leftrightarrow a^3+b^3+c^3=-3ab\left(a+b\right)\)
\(\Leftrightarrow a^3+b^3+c^3=-3ab.\left(-c\right)\)
\(\Leftrightarrow a^3+b^3+c^3=3abc\)
=> Điều phải chứng minh
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