Algebra

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! :)