Ta có : a^3+b^3+c^3=(a+b+c).(a^2+b^2+c^2-a.b-b.c-a.c)+3.a.b.c=3.a.b.c
=(a+b+c).(a^2+b^2+c^2-a.b-b.c-a.c)=0
Ta thấy:a,b,c là số dương nên a+b+c khác 0 suy ra (a^2+b^2+c^2-a.b-b.c-a.c) =0 nên a=b=c
Vậy a=b=c
\(a^3+b^3+c^3=3abc\)
\(\Leftrightarrow a^3+b^3+c^3-3abc=0\)
\(\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=0\)
\(\Leftrightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+2ab+b^2+ac+bc+c^2-3ab\right)=0\)
\(\Leftrightarrow a^2+b^2+c^2-ab-bc-ca=0\left(a+b+c>0\right)\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ac=0\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ac+a^2\right)=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
\(\Rightarrow\hept{\begin{cases}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\\left(c-a\right)^2=0\end{cases}\Rightarrow a=b=c}\)
#Thang Tran
Từ a3+b3+c3 =3abc suy ra a=b=c
Chứ không phải a=b=c suy ra a3+b3+c3 =3abc
a = b = c thay vào ta có :
a^3 + b^3 + c^3 = a^3 + a^3 + a^3 = 3a^3 (1)
3abc = 3.a.a.a = 3 a^3 (2)
Từ (1) và (2) => a^3 + b^3 + c^3 = 3abc
VẬy a = b= c thì ........
Ta có:
\(a^3+b^3+c^3=3abc\Leftrightarrow a^3+b^3+c^3-3abc=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca=0\right)\)
\(\Leftrightarrow a^2+b^2+c^2-ab-bc-ca=0\text{ (do }a+b+c>0\text{)}\)
\(\Leftrightarrow\frac{1}{2}\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]=0\)
\(\Leftrightarrow a-b=b-c=c-a=0\)
\(\Leftrightarrow a=b=c\)
Ta có :
-, \(a^3+b^3+c^3-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)
Do đó nếu \(a^3+b^3+c^3=3abc\) và \(a,b,c>0\)thì
\(a^2+b^2+c^2-ab-bc-ca=0\)
=>\(a^2+b^2+c^2=ab+bc+ca\)
=>\(2a^2+2b^2+2c^2=2ab+2bc+2ca\)
=>\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
=>\(a=b=c\)
