## Task A-1.2.

Let $x$ and $y$ be real numbers such that $x+y=1$ and $x^{3}+y^{3}=13$. What is $x^{2}+y^{2}$?
## First Solution.

From the formula for the cube of a sum

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

it follows that

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

By substituting $x+y=1$ and $x^{3}+y^{3}=13$, we have

$$
3 x y=1^{3}-13=-12
$$

which means $x y=-4$.

Now, from the formula for the square of a sum $(x+y)^{2}=x^{2}+2 x y+y^{2}$, by substituting $x+y=1$ and $x y=-4$, we have

$$
1^{2}=x^{2}+y^{2}+2 \cdot(-4)=x^{2}+y^{2}-8
$$

which means $x^{2}+y^{2}=1+8=9$.