## Task B-1.1.

The sum of two numbers is 6. If the sum of their cubes is 90, what is the sum of their squares?
## First Solution.

Let the required numbers be $x$ and $y$. According to the conditions of the problem, the following equations hold:

$x+y=6$ and $x^{3}+y^{3}=90$.

By applying the formula for the cube of a binomial and the above equations, we get:

| $(x+y)^{3}=x^{3}+3 x^{2} y+3 x y^{2}+y^{3}$ | $1 \text{point}$ |
| :--- | :--- |
| $(x+y)^{3}=x^{3}+y^{3}+3 x y(x+y)$ | $1 \text{point}$ |
| $6^{3}=90+3 x y \cdot 6$ | $1 \text{point}$ |
| $18 x y=126$ | $1 \text{point}$ |

Now, using the formula for the square of a binomial, express the required sum of squares:

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

Thus, $x^{2}+y^{2}=6^{2}-2 \cdot 7=22$.

1 point

Note:

The equality $x y=7$ can also be obtained by applying the formula for the sum of cubes:

$$
\begin{array}{lc}
x^{3}+y^{3}=(x+y)\left(x^{2}-x y+y^{2}\right) & 1 \text{ point } \\
90=6 \cdot\left(x^{2}-x y+y^{2}\right) & \\
15=x^{2}-x y+y^{2} & 1 \text{ point } \\
15=(x+y)^{2}-3 x y & 1 \text{ point } \\
15=6^{2}-3 x y & \\
3 x y=21 & 1 \text{ point }
\end{array}
$$