Fun with Surds

$a-c=(a-b) +(b-c) =3+\sqrt 8+3-\sqrt 8=6$

$a^2+b^2+c^2-ab-bc-ac=\dfrac 12 \left ((a-b) ^2+(b-c) ^2+(a-c) ^2\right )$

$=\dfrac 12 \left ((3+\sqrt 8)^2+(3-\sqrt 8)^2+6^2\right )$

$=\dfrac 12 \left (2(3^2+(\sqrt 8)^2)+36\right )=17+18=\boxed {35}$ .

$\begin{aligned} (a-b)^2 + (b-c)^2 + (a-b)(b-c) & = (3+\sqrt 8)^2 + (3-\sqrt 8)^2 + (3+\sqrt 8)(3+\sqrt 8) \\ a^2 - 2ab + b^2 + b*2 - 2bc + c^2 + ab - ca - b^2 + bc & = 17 + 2\sqrt 8 + 17 - 2\sqrt 8 + 9 - 8 \\ \implies a^2 + b^2 + c^2 - ab - bc - ca & = \boxed {35} \end{aligned}$

Squaring the two given equations gives us most of the terms we want, except $ac$ .

So first, adding the equations, $a-c=6$

Now squaring and summing, $(a-b)^2+(b-c)^2+(a-c)^2=(3+\sqrt8)^2+(3-\sqrt8)^2+6^2=70$

And also $(a-b)^2+(b-c)^2+(a-c)^2=2\left(a^2+b^2+c^2-ab-bc-ac\right)$

So $a^2+b^2+c^2-ab-bc-ac=\boxed{35}$

An easy triple to check this with is $(a,b,c)=(3,-\sqrt8,-3)$ .

First observe that $(a - b)(b - c) = 9 - 8 = 1 \rightarrow ab - b^{2} - ac + bc$

Secondly, $a - c = 6$ , and the required expression can be written as $(a - c)^{2} - (ab - b^{2} - ac + bc) = 36 - 1 = \boxed{35}$