Đáp án: $A=\dfrac{a}{b}$ hoặc $A=4$
Giải thích các bước giải:
Ta có:
$a^3+b^3+c^3-3abc=(a+b)^3-3ab(a+b)+c^3-3abc$
$\to a^3+b^3+c^3-3abc=(a+b)^3+c^3-(3ab(a+b)+3abc)$
$\to a^3+b^3+c^3-3abc=(a+b+c)^3-3(a+b)c(a+b+c)-3ab(a+b+c)$
$\to a^3+b^3+c^3-3abc=(a+b+c)^3-3(ac+bc)(a+b+c)-3ab(a+b+c)$
$\to a^3+b^3+c^3-3abc=(a+b+c)^3-3(ac+bc+ab)(a+b+c)$
$\to a^3+b^3+c^3-3abc=(a+b+c)((a+b+c)^2-3(ac+bc+ab))$
Mà $a^3+b^3+c^3=3abc\to a^3+b^3+c^3-3abc=0$
$\to (a+b+c)((a+b+c)^2-3(ac+bc+ab))=0$
$\to a+b+c=0$ hoặc $(a+b+c)^2-3(ac+bc+ab)=0$
Nếu $a+b+c=0$
$\to A=(1+\dfrac{a}{b})(1+\dfrac{b}{c})$
$\to A=\dfrac{a+b}{b}.\dfrac{b+c}{c}$
$\to A=\dfrac{-c}{b}.\dfrac{-a}{c}$
$\to A=\dfrac{a}{b}$
Nếu $(a+b+c)^2-3(ab+bc+ca)=0\to (a+b+c)^2=3(ab+bc+ca)$
Mà $(a+b+c)^2\ge 3(ab+bc+ca)$
$\to $Dấu = xảy ra khi $a=b=c$
$\to A=(1+1)(1+1)=4$
