Difference of cubic vieta's

Using Vieta's formula ,

$a+b+c=\dfrac{1}{2} \\ ab+bc+ac = \dfrac{1}{2}$

Using Newton's identity ,

$a^2+b^2+c^2 = (a+b+c)^2 - 2(ab+bc+ac) = \left(\dfrac{1}{2}\right)^2 - 2\left(\dfrac{1}{2}\right) = \dfrac{-3}{4}$

By using identity $x^3-y^3=(x-y)(x^2+xy+y^2)$ ,the terms like $(a-b) , (b-c),(c-a)$ get cancelled $(a \neq b \neq c)$ and thus we can write the given expression as :

$a^2+ab+b^2+b^2+bc+c^2+c^2+ac+a^2 \\ = 2(a^2+b^2+c^2)+ab+bc+ac \\ = 2\left(\dfrac{-3}{4}\right) + \dfrac{1}{2} \\ =\Large\boxed{-1}$

Simplify the given expression to get $2(a^2 + b^2 + c^2) + ab +bc+ac$ . Since $a^2+b^2+c^2 = (a+b+c)^2 - 2(ab+bc+ac)$ , we have

$2(a^2 + b^2 + c^2) + ab +bc+ac$

$= 2[(a+b+c)^2 - 2(ab+bc+ac)] + ab+bc+ac$

$=2(a+b+c)^2 - 4(ab+bc+ac)+ab+bc+ac$

$=2(a+b+c)^2 - 3(ab+bc+ac)$ . Substitute in the values and we are done.

Cancelling out from numerator and denominator common factors, the expression reduces to $2\sum {a}^2 + \sum ab$ , where $\sum ab = 1/2$ and $\sum{a}^2 = -3/4$ from the given cubic equation, (using $[\sum a]^2 = \sum {a}^2 + 2\sum ab$ ) to get the answer = -1.