$K=\left( 1+\frac{a}{2b} \right).\left( 1+\frac{2b}{3c} \right).\left( 1+\frac{3c}{a} \right)=\left( \frac{a+2b}{2b} \right).\left( \frac{2b+3c}{3c} \right).\left( \frac{3c+a}{a} \right)$
Ta có : ${{a}^{3}}+8{{b}^{3}}+27{{c}^{3}}=18abc$
$\Leftrightarrow ({{a}^{3}}+8{{b}^{3}}+3{{a}^{2}}.2b+3a.4{{b}^{2}})+27{{c}^{3}}-(3{{a}^{2}}.2b+3a.4{{b}^{2}})=18abc$
$\Leftrightarrow {{(a+2b)}^{3}}+27{{c}^{3}}-(6{{a}^{2}}b+12a{{b}^{2}})-18abc=0$
$\Leftrightarrow (a+2b+3c)\left[ {{(a+2b)}^{2}}-(a+2b).3c+9{{c}^{2}} \right]-(6{{a}^{2}}b+12a{{b}^{2}}+18abc)=0$
$\Leftrightarrow (a+2b+3c)({{a}^{2}}+4ab+4{{b}^{2}}-3ac-6bc+9{{c}^{2}})-6ab(a+2b+3c)=0$
$\Leftrightarrow (a+2b+3c)({{a}^{2}}+4ab+4{{b}^{2}}-3ac-6bc+9{{c}^{2}}-6ab)=0$
$\Leftrightarrow (a+2b+3c)({{a}^{2}}+4{{b}^{2}}+9{{c}^{2}}-3ac-6bc-2ab)=0$
$\Leftrightarrow \left[ \begin{align}& a+2b+3c=0 \\ & {{a}^{2}}+4{{b}^{2}}+9{{c}^{2}}-3ac-6bc-2ab=0 \\
\end{align} \right.$
Trường hợp 1: $a+2b+3c=0$$\Rightarrow \left\{ \begin{align}& a+2b=-3c \\ & 2b+3c=-a \\ & 3c+a=-2b \\ \end{align} \right.$

$K=\left( 1+\frac{a}{2b} \right).\left( 1+\frac{2b}{3c} \right).\left( 1+\frac{3c}{a} \right)=\left( \frac{a+2b}{2b} \right).\left( \frac{2b+3c}{3c} \right).\left( \frac{3c+a}{a} \right)$
$=\frac{-3c}{2b}.\frac{-a}{3c}.\frac{-2b}{a}=\frac{-6abc}{6abc}=-1$
Trường hợp 2: ${{a}^{2}}+4{{b}^{2}}+9{{c}^{2}}-3ac-6bc-2ab=0$
$\begin{align}& \Leftrightarrow 2{{a}^{2}}+8{{b}^{2}}+18{{c}^{2}}-6ac-12bc-4ab=0 \\
& \Leftrightarrow ({{a}^{2}}-4ab+4{{b}^{2}})+(4{{b}^{2}}-12bc+9{{c}^{2}})+({{a}^{2}}-6ac+9{{c}^{2}})=0 \\
& \Leftrightarrow {{(a-2b)}^{2}}+{{(2b-3c)}^{2}}+{{(a-3c)}^{2}}=0 \\ \end{align}$
Vì ${{(a-2b)}^{2}}\ge 0,\,\,{{(2b-3c)}^{2}}\ge 0,\,\,\,{{(a-3c)}^{2}}\ge 0\,\,\,\forall a,b,c$
$\Rightarrow \left\{ \begin{align} & {{(a-2b)}^{2}}=0 \\ & {{(2b-3c)}^{2}}=0 \\ & {{(a-3c)}^{2}}=0 \\ \end{align} \right.$$\Leftrightarrow \left\{ \begin{align}& a-2b=0 \\ & 2b-3c=0 \\ & a-3c=0 \\ \end{align} \right.$
$K=\left( 1+\frac{a}{2b} \right).\left( 1+\frac{2b}{3c} \right).\left( 1+\frac{3c}{a} \right)=\left( \frac{a+2b}{2b} \right).\left( \frac{2b+3c}{3c} \right).\left( \frac{3c+a}{a} \right)$
$=\frac{4b}{2b}.\frac{6c}{3c}.\frac{2a}{a}=\frac{48abc}{6abc}=8$