Bài 1
a³+b³+c³ = 3abc⇒a³+b³+c³ − 3abc=0
=> a = b = c
Và a + b + c = 0
Còn bài 2 gửi sau nha
Ta có: \(a^3+b^3+c^3-3abc=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)
Suy ra \(\frac{1}{2}\left(a+b+c\right)\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]=0\)
Suy ra \(a+b+c=0\) hoặc a = b = c.
b) Với a + b + c = 0 thì \(n=\frac{abc}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}=\frac{abc}{-abc}=-1\)
Với a = b = c thì \(n=\frac{a^3}{\left(2a\right)^3}=\frac{a^3}{8a^3}=\frac{1}{8}\)
Vậy ....
Chứng minh HĐT \(a^3+b^3+c^3-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\) nè!
\(VT=\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc\)
\(=\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=VP\)
