The answer is 1. We can factor $a^3+b^3$ as $(a+b)(a^2-ab+b^2) = (1)((a+b)^2-3ab) = (1-3q).$ We can factor $3(a^3b+ab^3)$ as $3ab(a^2+b^2) = 3q((a+b)^2-2ab) = 3q(1-2q).$ We can factor $6(a^3b^2+a^2b^3)$ as $6a^2b^2(a+b) = 6q^2.$ Combining, we find that $a^3+b^3+3(a^3b+ab^3)+6(a^3b^2+a^2b^3) = 1-3q+3q(1-2q)+6q^2 = 1-3q+3q-6q^2+6q^2 = 1$ Hope this helps! Please tell me if there are any errors with my solution.

Since $a,b$ are the roots of the quadratic equation, we have: $a+b=1 \wedge ab=1$ $a^3+b^3=(a+b)(a^2-ab+b^2)=1 \cdot (a^2+b^2-1)=(a+b)^2-2ab-1=1-2-1=-2$ $3(a^3b+ab^3)=3ab(a^2+b^2)=3 \cdot (-1)=-3$ $6(a^3b^2+a^2b^3)=6a^2b^2(a+b)=6(ab)^2 \cdot 1=6 \cdot 1=6$ $\Rightarrow a^3+b^3+3(a^3b+ab^3)+6(a^3b^2+a^2b^3)=-2-3+6=1$ $\blacksquare$