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$\sqrt3 +1 \;\text{and}\; -\sqrt3+1 \;\text{are two roots of this equation}$

so $3( \sqrt3 +1 )^3+a(\sqrt3 +1)^2+b(\sqrt3 +1)+12 =0$

$3( 3\sqrt3 +9+3\sqrt3+1 )+a(3+2\sqrt3 +1)+b(\sqrt3 +1)+12 =0$

$a(3+2\sqrt3 +1)+b(\sqrt3 +1)+ 18\sqrt3+42 =0$        

$a(4+2\sqrt3)+b(\sqrt3 +1)+ 18\sqrt3+42 =0$                                   -----(1)

using other root,

$3(-\sqrt3+1 )^3+a(-\sqrt3+1 )^2+b(-\sqrt3+1 )+12=0$

$3(-3\sqrt3+9-3\sqrt3+1 )+a(3-2\sqrt3+1 )+b(-\sqrt3+1 )+12=0$

$a(3-2\sqrt3+1 )+b(-\sqrt3+1 )-18\sqrt3+42=0$       

$a(4-2\sqrt3)+b(-\sqrt3+1 )-18\sqrt3+42=0$                               -------(2)

using (1) and (2)  

$4a+b=-42$                   [ by adding ]

$2a+b=-18$                    [ by subtracting]

$a=-12\; \text{and}\;b=6$

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