Algebra

We basically have all the information needed to solve this problem. 

First off, let's note that 

\((x+y)^3 = x^{3}+3x^{2}y+3xy^{2}+y^{3}\)

Simplifying this, we get

\(8=5+6xy\\ xy=1/2\)

Now, let's note that \((x+y)^2 = x^2+2xy+y^2\)

This contains x^2+y^2, so let's isolate that. We get

\(x^2+y^2 = (x+y)^2 - 2xy\)

We already have all the values shown for us to find x^2+y^2. Plugging in 2 and 1/2, we get

\(x^2 + y^2 = 4 - 1 = 3\)

So 3 is our final answer. 

Thanks! :)

x^3 + y^3 = (x + y) (x^2 + y^2  - xy)

        5 =  (2) (x^2 + y^2 - xy)

       (5/2)  = x^2 + y^2  - xy

      (5/2) + xy  = x^2 + y^2

x + y = 2      square both sides

x^2 + y^2  + 2xy  = 4

x^2 + y^2  = 4  - 2xy

So

(5/2) + xy  =  4 - 2xy

3xy  =  4 - 5/2

3xy =  3/2

xy =  1/2

2xy= 1   

So

x^2 + y^2  = 4 - 1  =   3