Bài giải:

${{(a-b)}^{2}}+{{(b-c)}^{2}}+{{(a-c)}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}$

${{a}^{2}}-2ab+{{b}^{2}}+{{b}^{2}}-2bc+{{c}^{2}}+{{a}^{2}}-2ac+{{c}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}$

$2({{a}^{2}}+{{b}^{2}}+{{c}^{2}})-2ab-2bc-2ac={{a}^{2}}+{{b}^{2}}+{{c}^{2}}$

$\Rightarrow $ ${{a}^{2}}+{{b}^{2}}+{{c}^{2}}=2ab+2bc+2ac$            (1)

Mà: a +b + c = 6 nên ${{(a+b+c)}^{2}}=36$

$\Rightarrow $${{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2ab+2bc+2ac=36$     (2)

Từ (1) và (2) suy ra: $4(ab+bc+ac)=36$

$\Rightarrow $ab + bc + ac = 9

Vậy ab + bc + ac = 9