Đáp án: $\{-1,8\}$
Giải thích các bước giải:
Ta có:
$a^3+b^3-c^3=-3abc$
$\to a^3+b^3-c^3+3abc=0$
$\to (a+b)^3-3ab(a+b)-c^3+3abc=0$
$\to (a+b)^3-c^3-3ab(a+b)+3abc=0$
$\to (a+b-c)((a+b)^2+(a+b)c+c^2)-3ab(a+b-c)=0$
$\to (a+b-c)((a+b)^2+(a+b)c+c^2-3ab)=0$
$\to (a+b-c)(a^2+b^2+c^2-ab+bc+ca)=0$
$\to a+b-c=0$ hoặc $a^2+b^2+c^2-ab+bc+ca=0$
Giải $a+b-c=0$
$\to a+b=c, a-c=-b, c-b=a$
Ta có:
$(1+\dfrac ab)(1-\dfrac bc)(1-\dfrac ca)=\dfrac{a+b}{b}\cdot \dfrac{c-b}{c}\cdot \dfrac{a-c}{a}=\dfrac{c}{b}\cdot \dfrac{a}c\cdot \dfrac{-b}{a}=-1$
Giải $a^2+b^2+c^2-ab+bc+ca=0$
$\to 2(a^2+b^2+c^2-ab+bc+ca)=0$
$\to (a^2-2ab+b^2)+(b^2+2bc+c^2)+(c^2+2ca+a^2)=0$
$\to (a-b)^2+(b+c)^2+(c+a)^2=0$
Mà $(a-b)^2+(b+c)^2+(c+a)^2\ge 0$
$\to$Dấu = xảy ra khi $(a-b)^2=(b+c)^2=(c+a)^2=0$
$\to a-b=b+c=c+a=0$
$\to a=b, b=-c, c=-a$
$\to a=b=-c$
$\to (1+\dfrac ab)(1-\dfrac bc)(1-\dfrac ca)=(1+\dfrac aa)(1-\dfrac{-c}{c})(1-\dfrac{c}{-c})=8$
