jmc

Let $r$ and $s$ be the two common roots. Then $r$ and $s$ are the roots of $$(x^3+a{x}^2+11x+6)-(x^3+b{x}^2+14+8)=(a-b)x^2-3x-2.$$Note that $r$ and $s$ are also the roots of $$\begin{array}{rl}& 4(x^3+a{x}^2+11x+6)-3(x^3+b{x}^2+14x+8)\\ & =x^3+(4a-3b)x^2+2x\\ & =x[x^2+(4a-3b)x+2].\end{array}$$Since the constant coefficient of $x^3+a{x}^2+11x+6$ is nonzero, both $r$ and $s$ are non-zero. Therefore, $r$ and $s$ are the roots of $$x^2+(4a-3b)x+2.$$Hence, both $r$ and $s$ are the roots of $-x^2+(3b-4a)x-2.$ But $r$ and $s$ are also the roots of $(a-b)x^2-3x-2,$ so the coefficients must match. This gives us $a-b=-1$ and $3b-4a=-3.$ Solving, we find $(a,b)=\boxed{(6,7)}.$

For these values, the given cubics become $$\begin{array}{rl}{x}^3+6x^2+11x+6& =(x+1)(x+2)(x+3),\\ x^3+7x^2+14x+8& =(x+1)(x+2)(x+4).\end{array}$$