a3+b3+c3=3abc
⇔a3+b3+c3−3abc=0
⇔(a+b)3+c3−3ab(a+b)−3abc=0
⇔(a+b+c)[(a+b)2−(a+b)c+c2]−3ab(a+b+c)=0⇔(a+b+c)(a2+b2+c2−ab−bc−ca)=0
⇔12(a+b+c)(2a2+2b2+2c2−2ab−2bc−2ac)=0
⇔12(a+b+c)[(a−b)2+(b−c)2+(c−a)2]=0
Vì a,b,c là cạnh của các tam giác nên a,b,c>0
=>\(\left\{{}\begin{matrix}a-b=0\\b-c=0\\c-a=0\end{matrix}\right.\)
=>a=b=c
Vậy tam giác đó đều
Vi
a3+b3+c3=3abc
=>a3+b3+c3-3abc=0
=>a3+3a2b+3ab2+b3+c3-3abc-3a2b-3ab2=0(thêm bớt 3a2b và 3ab2)
=>(a3+3a2b+3ab2+b3)+c3-(3a2b+3ab2+3abc)=0
=>(a+b)3+c3-3ab(a+b+c)=0
=>(a+b+c)[(a+b)2-(a+b)+c+c2]-3ab(a+b+c)=0
=>(a+b+c)(a2+2ab+b2-ac-bc+c2-3ab)=0
=>a+b+c=0
mà a;b;c>0
=>a,b,c=0
=>a=b=c
vậy .....
Ta có:
\(a^3+b^3+c^3=3abc\)
\(\Leftrightarrow\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc=0\)
\(\Leftrightarrow\left(a+b\right)^3-3ab\left(a+b+c\right)+c^3=0\)
\(\Leftrightarrow\left(a+b+c\right)^3-3ab\left(a+b+c\right)-3c\left(a+b\right)\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)^3-3\left(a+b+c\right)\left(ab+bc+ca\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2+2ab+2bc+2ca-3ab-3ac-3bc\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)
Mà a,b,c \(>0\Rightarrow a+b+c>0\)
Nên \(\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)
\(2\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)
\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
=> a-b=0
b-c=0 => a=b=c => tam giác đó là tam giác đều
c-a=0
