Cách 1 . \(A=\left(\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b}\right)\left(\frac{c}{a-b}+\frac{a}{b-c}+\frac{b}{c-a}\right)\)
Đặt \(\frac{a-b}{c}=x\); \(\frac{b-c}{a}=y\) ; \(\frac{c-a}{b}=z\)
Ta có : \(\frac{x+y}{z}=\frac{\frac{a-b}{c}+\frac{b-c}{a}}{\frac{c-a}{b}}=\frac{ab\left(a-b\right)+cb\left(b-c\right)}{ac\left(c-a\right)}=\frac{b\left(b-a-c\right)}{ac}=\frac{2b^2}{ac}=\frac{2b^3}{abc}\)
tương tự : \(\frac{y+z}{x}=\frac{2c^3}{abc}\); \(\frac{x+z}{y}=\frac{2a^3}{abc}\)
\(\Rightarrow A=\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=1+\frac{x}{y}+\frac{x}{z}+\frac{y}{x}+1+\frac{y}{z}+\frac{z}{x}+\frac{z}{y}+1\)
\(=3+\frac{x+z}{y}+\frac{x+y}{z}+\frac{y+z}{x}=3+\frac{2}{abc}\left(a^3+b^3+c^3\right)\)
Áp dụng bài toán phụ : Nếu a + b + c = 0 thì \(a^3+b^3+c^3=3abc\) (có thể chứng minh bằng cách rút a = - b - c rồi thay vào tổng ba lập phương) được :
\(A=3+\frac{2}{abc}.3abc=3+6=9\)
Đặt \(\frac{a-b}{c}=x=>\frac{c}{a-b}=\frac{1}{x}\)
\(\frac{b-c}{a}=y=>\frac{a}{b-c}=y\)
\(\frac{c-a}{b}=z=>\frac{b}{c-a}=\frac{1}{z}\)
=>\(A=\left(x+y+z\right).\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)
=>\(A=x.\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)+y.\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)+z.\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)
=>\(A=1+\frac{x}{y}+\frac{x}{z}+1+\frac{y}{x}+\frac{y}{z}+1+\frac{z}{x}+\frac{z}{y}\)
=>\(A=3+\left(\frac{x}{y}+\frac{x}{z}+\frac{y}{x}+\frac{y}{z}+\frac{z}{x}+\frac{z}{y}\right)\)
=>\(A=3+\frac{x+z}{y}+\frac{x+y}{z}+\frac{y+z}{x}\)
Lại có: \(\frac{x+z}{y}=\frac{\frac{a-b}{c}+\frac{c-a}{b}}{\frac{b-c}{a}}=\frac{\frac{ab-b^2}{bc}+\frac{c^2-ac}{bc}}{\frac{b-c}{a}}=\frac{\frac{ab-b^2+c^2-ac}{bc}}{\frac{b-c}{a}}\)
\(=\frac{\frac{\left(ab-ac\right)-\left(b^2-c^2\right)}{bc}}{\frac{b-c}{a}}=\frac{\frac{a.\left(b-c\right)-\left(b+c\right).\left(b-c\right)}{bc}}{\frac{b-c}{a}}=\frac{\frac{\left(a-b-c\right).\left(b-c\right)}{bc}}{\frac{b-c}{a}}\)
\(=\frac{\left(a-b-c\right).\left(b-c\right).a}{\left(b-c\right).bc}=\frac{\left(a-b-c\right).a}{bc}=\frac{\left(a+a-a-b-c\right).a}{bc}\)
\(=\frac{\left[2a-\left(a+b+c\right)\right].a}{bc}\)
Vì a+b+c=0
=>\(\frac{x+z}{y}=\frac{\left(2a-0\right).a}{bc}=\frac{2a^2}{bc}=\frac{2a^3}{abc}\)
Chứng minh tương tự, ta có:
\(\frac{x+y}{z}=\frac{2b^3}{abc}\)
\(\frac{y+z}{x}=\frac{2c^3}{abc}\)
=>\(A=3+\frac{x+z}{y}+\frac{x+y}{z}+\frac{y+z}{x}=3+\frac{3a^3}{abc}+\frac{3b^3}{abc}+\frac{3c^3}{abc}\)
=>\(A=3+\frac{2a^3+2b^3+2c^3}{abc}\)
=>\(A=3+\frac{2.\left(a^3+b^3+c^3\right)}{abc}\)
Vì a+b+c=0
=>a=-(b+c)
=>\(a^3=\left[-\left(b+c\right)\right]^3\)
=>\(a^3=-\left(b+c\right)^3\)
=>\(a^3=-\left[b^3+3bc.\left(b+c\right)+c^3\right]\)
=>\(a^3=-b^3-3bc.\left(b+c\right)-c^3\)
=>\(a^3+b^3+c^3=-3bc.\left(b+c\right)\)
Vì a+b+c=0=>b+c=-a
=>\(a^3+b^3+c^3=-3bc.\left(-a\right)\)
=>\(a^3+b^3+c^3=3abc\)
Thay vào A, ta có:
\(A=3+\frac{2.\left(a^3+b^3+c^3\right)}{abc}=3+\frac{2.3abc}{abc}=3+\frac{6.abc}{abc}=3+6=9\)
=>A=9
Vậy A=9
Cách 2. Đặt \(P=\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b}\) ; \(Q=\frac{c}{a-b}+\frac{a}{b-c}+\frac{b}{c-a}\)
\(\Rightarrow P=\frac{ab\left(a-b\right)+bc\left(b-c\right)+ac\left(c-a\right)}{abc}\)
Xét riêng : \(ab\left(a-b\right)+bc\left(b-c\right)+ac\left(c-a\right)=ab\left[-\left(b-c\right)-\left(c-a\right)\right]+bc\left(b-c\right)+ac\left(c-a\right)\)
\(=\left[-ab\left(b-c\right)+bc\left(b-c\right)\right]+\left[-ab\left(c-a\right)+ac\left(c-a\right)\right]\)
\(=b.\left(c-a\right).\left(b-c\right)+a\left(c-a\right)\left(c-b\right)=\left(c-a\right)\left(b-c\right)\left(b-a\right)\)
Vậy : \(P=\frac{\left(c-a\right)\left(b-c\right)\left(b-a\right)}{abc}\)
Tiếp theo, rút gọn Q như sau :
Đặt \(x=b-c\); \(y=c-a\); \(z=a-b\)
Ta có : \(x-y=a+b-2c=-c-2c=-3c\)
\(y-z=b+c-2a=-a-2a=-3a\)
\(z-x=c+a-2b=-b-2b=-3b\)
\(\Rightarrow3Q=\frac{-\left(y-z\right)}{x}+\frac{-\left(z-x\right)}{y}+\frac{-\left(x-y\right)}{z}\)\(\Rightarrow-3Q=\frac{y-z}{x}+\frac{z-x}{y}+\frac{x-y}{z}\)
Rút gọn tương tự như P, ta được : \(-3Q=\frac{\left(x-y\right)\left(y-z\right)\left(x-z\right)}{xyz}=\frac{\left(-3c\right).\left(-3a\right).\left(3b\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(\Rightarrow Q=-\frac{9abc}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
Vậy : \(A=PQ=\frac{\left(c-a\right)\left(c-b\right)\left(a-b\right)}{abc}.\frac{-9abc}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)\(\Rightarrow A=9\)
