Question

[Zero Two]

Cho a . b . c đối một khác nhau thỏa mãn a³ + b³ + c³ = 3abc

Tính M = (a + b) (b + c) (c + a)

Answer 1

[Nobi Nobita]

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

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

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

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

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

\(\Leftrightarrow\left(a+b+c\right)\left[\left(a+b+c\right)^2-3\left(a+b\right).c-3ab\right]=0\)

\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2+2ab+2bc+2ca-3ac-3bc-3ab\right)=0\)

\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)

\(\Leftrightarrow\frac{1}{2}\left(a+b+c\right)\left(2a^2+2b^2+2c^2-2ab-2bc-2ca\right)=0\)

\(\Leftrightarrow\frac{1}{2}\left(a+b+c\right)\left[\left(a^2-2ab+b^2\right)+\left(b^2-2bc+b^2\right)+\left(c^2-2ca+a^2\right)\right]=0\)

\(\Leftrightarrow\frac{1}{2}\left(a+b+c\right)\left[\left(a-b\right)^2+\left(c-a\right)^2+\left(c-a\right)^2\right]=0\)

\(\Leftrightarrow\orbr{\begin{cases}a+b+c=0\\\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\end{cases}}\)

TH1: Nếu \(a+b+c=0\)\(\Rightarrow\hept{\begin{cases}a+b=-c\\b+c=-a\\c+a=-b\end{cases}}\)

Thay vào biểu thức M ta có:

\(M=\left(-c\right).\left(-a\right).\left(-b\right)=-abc\)

TH2: Nếu \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)

Vì \(\left(a-b\right)^2\ge0\); \(\left(b-c\right)^2\ge0\); \(\left(c-a\right)^2\ge0\)với \(\forall a,b,c\)

\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)

Dấu " = " xảy ra \(\Leftrightarrow\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}}\Leftrightarrow a=b=c\)

Thay vào biểu thức M ta có:

\(M=\left(a+b\right)\left(b+c\right)\left(c+a\right)=2a.2b.2c=8abc\)

Vậy \(M=-abc\)hoặc \(M=8abc\)