The answer is 1.
We can factor a3+b3 as (a+b)(a2−ab+b2)=(1)((a+b)2−3ab)=(1−3q).
We can factor 3(a3b+ab3) as 3ab(a2+b2)=3q((a+b)2−2ab)=3q(1−2q).
We can factor 6(a3b2+a2b3) as 6a2b2(a+b)=6q2.
Combining, we find that a3+b3+3(a3b+ab3)+6(a3b2+a2b3)=1−3q+3q(1−2q)+6q2=1−3q+3q−6q2+6q2=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∧ab=1a3+b3=(a+b)(a2−ab+b2)=1⋅(a2+b2−1)=(a+b)2−2ab−1=1−2−1=−23(a3b+ab3)=3ab(a2+b2)=3⋅(−1)=−36(a3b2+a2b3)=6a2b2(a+b)=6(ab)2⋅1=6⋅1=6⇒a3+b3+3(a3b+ab3)+6(a3b2+a2b3)=−2−3+6=1◼