## Task B-1.1.

For real numbers $x$ and $y$, if $x-y=6$ and $x^{2}+y^{2}=22$, what is $x^{3}-y^{3}$?
## Solution.

We have

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

Now, substitute $x-y$ with 6, and $x^{2}+y^{2}$ with 22, and we get

$$
x^{3}-y^{3}=6(x y+22)
$$

It remains to calculate the product $x y$.

After squaring the equation $x-y=6$, we get

$$
\begin{array}{cc}
(x-y)^{2}=36 & 1 \text { point } \\
x^{2}-2 x y+y^{2}=36 & \\
22-2 x y=36 & 1 \text { point } \\
x y=-7 & 1 \text { point }
\end{array}
$$

The value of the desired expression is

$$
x^{3}-y^{3}=6(x y+22)=90 \text {. }
$$