Cho a,b,c là 3 số khác 0 sao cho [tex]a^{3}b^{3} + b^{3}c^{3} + c^{3}a^{3} = 3a^{2}b^{2}c^{2}[/tex]
Tính M = [tex](1+\frac{a}{b})(1+\frac{b}{c})(1+\frac{c}{a})[/tex]
Nếu $x^3+y^3+z^3=3xyz$ thì $x+y+z=0$ hoặc $x=y=z$
Thật vậy, ta có:
$x^3+y^3+z^3=3xyz$
$\Leftrightarrow (x+y)^3+z^3-3xy(x+y)-3xyz=0$
$\Leftrightarrow (x+y+z)[(x+y)^2-z(x+y)+z^2]-3xy(x+y+z)=0$
$\Leftrightarrow (x+y+z)(x^2+y^2+z^2-xy-yz-zx)=0$
$\Leftrightarrow (x+y+z)(2x^2+2y^2+2z^2-2xy-2yz-2zx)=0$
$\Leftrightarrow (x+y+z)[(x^2-2xy+y^2)+(y^2-2yz+z^2)+(z^2-2zx+x^2)]=0$
$\Leftrightarrow (x+y+z)[(x-y)^2+(y-z)^2+(z-x)^2]=0$
$\Leftrightarrow x+y+z=0$ hoặc $x=y=z$
$(ab,bc,ca)\rightarrow (x,y,z)$
$\Rightarrow ab=bc=ca$ hoặc $ab+bc+ca=0$
Nếu $ab=bc=ca\Rightarrow a=b=c$ thì $M=(1+1)(1+1)(1+1)=8$
Nếu $ab+bc+ca=0\Rightarrow a+b=\dfrac{-ab}c;b+c=\dfrac{-bc}a;c+a=\dfrac{-ca}b$
$\Rightarrow M=\dfrac{(a+b)(b+c)(c+a)}{abc}=\dfrac{-abc}{abc}=-1$