Let $a$ and $b$ be complex numbers. If $a + b = 4$ and $a^2 + b^2 = 6,$ then what is $a^3 + b^3?$
Let's use a whole number of equations to solve this problem.
First off, we know that \((a+b)^2=a^2+2ab+b^2\)
We ALSO know that \((a+b)^2=4^2=16\)
Thus, we have the equation \(a^2+2ab+b^2=16\)
We are given a^2+b^2 in the problem, so plugging that in, we get
\(6+2ab=16\)
We can now find ab. This will come in handy later.
\(2ab=10\\ ab=5\)
Alright, let's move on to what we are TRYING to find.
Let's note that we can split a^3+b^3 into
\(a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)\)
WAIT! We already know the values of every number! Plugging in 4, 6, and 5, we get
We have
\((4)(6-5)=4\)
So 4 SHOULD be the answer.
I might have a mistake during the calculations...not sure.
Thanks! :)