Algebra

We don't actually have to find what a and b is. 

First, let's note that we have \((a+b)^2 = a^2+2ab+b^2\)

Squaring both sides of the first equation, we find that

\(a^2+2ab+b^2 = 16\)

Subtracting the second equation from this equation we just computed, we have

\(3ab = 10\\ ab = 10/3\)

Now, let's note something else real quick. 

We have that \((a+b)^3=a^{3}+3a^{2}b+3ab^{2}+b^{3} = a^3 + 3ab(a+b)+b^3\)

Since we also know that \((a+b)^3 = 4^3=64\), we have the equation \( a^3 + 3ab(a+b)+b^3=64\)

Wait! we already have all the numbers needed to isolate a^3 and b^3. We just found what ab was and a+b is given, so we have

\(a^3+b^3=64-3(10/3)(4) \\a^3+b^3 = 64-40\\ a^3+b^3=24\)

So 24 is our final answer. 

Thanks! :)

For this question, use formula: (a+b)^2 = a^2+b^2+2ab

since a+b=4

a^2+b^2+2ab = 16

a^2+b^2 = 6 + ab

substitute:

6 + ab + 2ab = 16

3ab + 6 =16

3ab = 10

ab= 10/3

a^3 + b^3 = (a+b)(a^2-ab+b^2)

a+b = 4, a^2+b^2 = 6 + 10/3 = 28/3 , and -ab = -10/3

28/3 - 10/3 = 6

=> 4*6 = 24

here's another way to understand :)

(good job, @NotThatSmart)