$E=\left( 1+\frac{a}{b} \right).\left( 1+\frac{b}{c} \right).\left( 1+\frac{c}{a} \right)=\frac{a+b}{b}.\frac{b+c}{c}.\frac{a+c}{a}$
Ta có : ${{a}^{3}}+{{b}^{3}}+{{c}^{3}}=3abc$
$\Leftrightarrow ({{a}^{3}}+{{b}^{3}}+3{{a}^{2}}b+3a{{b}^{2}})+{{c}^{3}}-(3{{a}^{2}}b+3a{{b}^{2}})=3abc$
$\Leftrightarrow {{(a+b)}^{3}}+{{c}^{3}}-(3{{a}^{2}}b+3a{{b}^{2}})-3abc=0$
$\Leftrightarrow (a+b+c)\left[ {{(a+b)}^{2}}-(a+b).c+{{c}^{2}} \right]-(3{{a}^{2}}b+3a{{b}^{2}}+3abc)=0$
$\Leftrightarrow (a+b+c)({{a}^{2}}+2ab+{{b}^{2}}-ac-bc+{{c}^{2}})-3ab(a+b+c)=0$
$\Leftrightarrow (a+b+c)({{a}^{2}}+2ab+{{b}^{2}}-ac-bc+{{c}^{2}}-3ab)=0$
$\Leftrightarrow (a+b+c)({{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ac-bc-ab)=0$
$\Leftrightarrow \left[ \begin{align}& a+b+c=0 \\ & {{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ac-bc-ab=0 \\ \end{align} \right.$

Trường hợp 1: $a+b+c=0$$\Rightarrow \left\{ \begin{align}& a+b=-c \\ & b+c=-a \\ & c+a=-b \\ \end{align} \right.$

$E=\left( 1+\frac{a}{b} \right).\left( 1+\frac{b}{c} \right).\left( 1+\frac{c}{a} \right)=\frac{a+b}{b}.\frac{b+c}{c}.\frac{a+c}{a}=\frac{-c}{b}.\frac{-a}{c}.\frac{-b}{a}=\frac{-abc}{abc}=-1$
Trường hợp 2: ${{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ac-bc-ab=0$
$\begin{align}& \Leftrightarrow 2{{a}^{2}}+2{{b}^{2}}+2{{c}^{2}}-2ac-2bc-2ab=0 \\ & \Leftrightarrow ({{a}^{2}}-2ab+{{b}^{2}})+({{b}^{2}}-2bc+{{c}^{2}})+({{a}^{2}}-2ac+{{c}^{2}})=0 \\ & \Leftrightarrow {{(a-b)}^{2}}+{{(b-c)}^{2}}+{{(a-c)}^{2}}=0 \\ \end{align}$
Vì ${{(a-b)}^{2}}\ge 0,\,\,{{(b-c)}^{2}}\ge 0,\,\,\,{{(a-c)}^{2}}\ge 0\,\,\,\forall a,b,c$

$E=\left( 1+\frac{a}{b} \right).\left( 1+\frac{b}{c} \right).\left( 1+\frac{c}{a} \right)=\frac{a+b}{b}.\frac{b+c}{c}.\frac{a+c}{a}=\frac{2b}{b}.\frac{2c}{c}.\frac{2a}{a}=\frac{8abc}{abc}=8$