+1459 Let $a$ and $b$ be complex numbers. If $a + b = 4$ and $a^2 + b^2 = 6 + 2ab,$ then what is $a^3 + b^3?$
+1926 First, look at the first equation. Squaring both sides, we have
\(a^2 + 2ab + b^2 = 16\)
Now, we have two equations to work with. We have
\(a^2 + b^2 = 16 - 2ab \\ a^2 + b^2 = 6 + 2ab \)
Subtracting the second equation from the first equation, we get
\(0 = -10 + 4ab \\ 10 = 4ab \\ ab = 10 / 4 = 5/2\)
Now, let's acknowledge something about a^3+b^3. Note that
\(a^3 + b^3 = (a + b) ( a^2 + b^2 - ab) \)
Wait! we already have all the terms needed to solve the problem!
We have
\( \\ (4) ( 6 + 2ab - ab) = \\ (4) ( 6 - ab) = \\ (4) ( 6 - 5/2) = \\ (4) ( 7/2) = \\ 14\)
So 14 is our answer.
Thanks! :)
