$\begin{align}  & {{x}^{3}}-(5+a){{x}^{2}}+(6+5a)x-6a=0 \\ & \text{Let }f(x)={{x}^{3}}-(5+a){{x}^{2}}+(6+5a)x-6a \\ & \text{Using trial and error method , let }x=a \\ & f(a)={{a}^{3}}-(5+a){{a}^{2}}+(6+5a)a-6a \\ & f(a)={{a}^{3}}-5{{a}^{2}}-{{a}^{3}}+6a+5{{a}^{2}}-6a \\ & f(a)=0 \\ & \therefore x-a\text{ is a factor of }f(x) \\ & \text{Expanding }f(x)\text{ will give } \\ & f(x)={{x}^{3}}-5{{x}^{2}}-a{{x}^{2}}+6x+5ax-6a \\ & \text{Using long division to find the other factors} \\ & x-a\overset{{{x}^{2}}-5x+6}{\overline{\left){\begin{align}  & {{x}^{3}}-5{{x}^{2}}-a{{x}^{2}}+6x+5ax-6a \\ & \underline{{{x}^{3}}\text{          }-a{{x}^{2}}\text{                          }} \\ & \text{      }-5{{x}^{2}}+6x+5ax-6a \\ & \text{  }\underline{\text{    }-5{{x}^{2}}\text{        }+5ax\text{         }} \\ & \text{                  }6x\text{            }-6a \\ & \text{ }\underline{\text{                 }6x\text{            }-6a}\text{      } \\\end{align}}\right.}} \\ & \text{                }------------ \\ & f(x)=({{x}^{2}}-5x+6)(x-a) \\ & f(x)=(x-3)(x-2)(x-a) \\ & \text{When }f(x)=0 \\ & (x-3)(x-2)(x-a)=0 \\ & x=3\text{ or }x=2\text{ or }x=a \\\end{align}$