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

It is ${x}^{2} + {y}^{2} = 3$

#### Explanation:

From $x + y = 2$ we have that

$x + y = 2 \implies {\left(x + y\right)}^{2} = {2}^{2} \implies {x}^{2} + {y}^{2} + 2 x y = 4$

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

${x}^{2} + {y}^{2} + 2 x y = 4$ (1)

and

${x}^{2} + {y}^{2} - x y = \frac{5}{2} \implies 2 \cdot \left({x}^{2} + {y}^{2}\right) - 2 x y = 5$ (2)

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

$3 \cdot \left({x}^{2} + {y}^{2}\right) = 9 \implies {x}^{2} + {y}^{2} = 3$

Dec 27, 2015

${x}^{2} + {y}^{2} = 3$

#### Explanation:

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

$5 = {x}^{3} + {y}^{3} = \left(x + y\right) \left({x}^{2} - x y + {y}^{2}\right) = 2 \left({x}^{2} + {y}^{2} - x y\right)$

$\implies {x}^{2} + {y}^{2} - x y = \frac{5}{2}$

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

Now, by cubing the first given equation, we get

${\left(x + y\right)}^{3} = {2}^{3}$

$\implies {x}^{3} + 3 {x}^{2} y + 3 x {y}^{2} + {y}^{3} = 8$

$\implies \left({x}^{3} + {y}^{3}\right) + 3 x y \left(x + y\right) = 8$

Substituting in our known values for $x + y$ and ${x}^{3} + {y}^{3}$, we obtain

$\implies 5 + 3 x y \cdot 2 = 8$

$\implies 6 x y = 3$

$\implies x y = \frac{1}{2}$

Substituting this back into $\left(\text{*}\right)$:

${x}^{2} + {y}^{2} = \frac{5}{2} + \frac{1}{2}$

$\therefore {x}^{2} + {y}^{2} = 3$