Question

Given `x+y=2` and `x^3+y^3=5`, what is `x^2+y^2`?

Answer 1

It is `x^2+y^2=3`

Explanation:

From `x+y=2` we have that

`x+y=2=>(x+y)^2=2^2=>x^2+y^2+2xy=4`

From `x^3+y^3=5` we have that

#x^3+y^3=5=>(x+y)*(x^2+y^2-xy)=5=> x^2+y^2-xy=5/2#

So we know that

`x^2+y^2+2xy=4` (1)

and

`x^2+y^2-xy=5/2=>2*(x^2+y^2)-2xy=5` (2)

If you add (1) and (2) you get

`3*(x^2+y^2)=9=>x^2+y^2=3`

Answer 2

`x^2 + y^2 = 3`

Explanation:

By applying the sum of cubes formula together with the given equations:

`5 = x^3 + y^3 = (x+y)(x^2-xy+y^2) = 2(x^2 + y^2 - xy)`

`=> x^2 + y^2 - xy= 5/2`

`=>x^2 + y^2 = 5/2 + xy" "("*")`

Now, by cubing the first given equation, we get

`(x+y)^3 = 2^3`

`=>x^3 + 3x^2y + 3xy^2 + y^3 = 8`

`=> (x^3 + y^3) + 3xy(x + y) = 8`

Substituting in our known values for `x+y` and `x^3 + y^3`, we obtain

`=> 5 + 3xy*2 = 8`

`=> 6xy = 3`

`=> xy = 1/2`

Substituting this back into `("*")`:

`x^2 + y^2 = 5/2 + 1/2`

`:. x^2 + y^2 = 3`