<PRE><B>Question 1079386</B>
Given: a+b=3, a^3+b^3=6
(a+b)^3 = a^3+3a^2b+3ab^2+b^3
Since a^3+b^3=6 and a+b=3, we have
3^3 = 6 + 3a^2b + 3ab^2
21 = 3a^2b + 3ab^2
Pulling out the common factor 3ab, we get
21 = 3ab(a+b) = 3*3*ab
Thus ab = 21/9 = 7/3
(a+b)^2 = a^2+b^2+2ab
Subtract 2ab from both sides:
(a+b)^2 - 2ab = a^2 + b^2
3^2 - 2(7/3) = a^2 + b^2
9 - 14/3 = (27-14)/3 = 13/3
So a^2 + b^2 = 13/3</PRE>