Ta có: ab+bc+ca=0
\(\Leftrightarrow\left\{{}\begin{matrix}ab+bc=-ca\\ab+ca=-bc\\bc+ca=-ab\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b\left(a+c\right)=-ca\\a\left(b+c\right)=-bc\\c\left(a+b\right)=-ab\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+c=\dfrac{-ca}{b}\\b+c=\dfrac{-bc}{a}\\a+b=\dfrac{-ab}{c}\end{matrix}\right.\)
Ta có: \(E=\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}\)
\(=\dfrac{-\dfrac{ab}{c}\cdot\dfrac{-bc}{a}\cdot\dfrac{-ca}{b}}{abc}\)
\(=\dfrac{-a^2b^2c^2}{abc}:abc\)
\(=\dfrac{-abc}{abc}=-1\)
Ta có: $(a+b)(b+c)(c+a)=(a+b)(a+c)(b+c)$
$=(a^2+ab+ac+bc)(b+c)$
$=[a(a+b+c)+bc].(b+c)$
$=a.(b+c)(a+b+c)+bc.(b+c)$
$=(a+b+c)(ab+ac)+bc(b+c)+abc-abc$
$=(a+b+c)(ab+ac)+bc.(a+b+c)-abc$
$=(a+b+c)(ab+bc+ca)-abc$
Mà $ab+bc+ca=0$
$⇒(a+b)(a+c)(b+c)=(a+b+c).0-abc=-abc$
Suy ra $E=\dfrac{(a+b)(a+c)(b+c)}{abc}=\dfrac{-abc}{abc}=-1$ (do $abc \neq 0$)
Vậy $E=-1$ với $a;b;c$ thỏa mãn đề
