Task B-2.7. Solutions of the equation

$$
\frac{(a-x)^{3}+(b-x)^{3}}{(a-x)^{2}+(b-x)^{2}}=a-b ; a, b, \in R
$$

are consecutive multiples of the number 6 whose sum of squares is 4068. Determine $a \cdot b$.
# Solution.

After simplifying the expression

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

we get

$$
(b-x)\left(2 x^{2}-x(a+3 b)+a^{2}+2 b^{2}-a b\right)=0
$$

We have several possibilities.

First possibility.

One solution to this equation is $x=b$.

If this is the only solution, then $b^{2}=4068$, and $b$ is not a natural number, nor a multiple of 6.

1 point

Second possibility.

One solution is $x=b=6 k$.

If the equation has two solutions, then the quadratic equation $2 x^{2}-x(a+3 b)+a^{2}+2 b^{2}-a b=0$ has one solution, so its discriminant is 0.

$$
D=0, \quad x=\frac{a+3 b}{4}=6 k \pm 6
$$

Now we have

$$
\begin{gathered}
(6 k)^{2}+(6 k \pm 6)^{2}=4068 \\
k^{2} \pm k=56, k \in N \\
k=7 \text { or } k=8
\end{gathered}
$$

For $k=7$, $b=x=42$ or $x=48$.

$b=42, a=66$.

1 point

For $k=8$, $b=x=48$ or $x=42$.

$b=48, a=24$.

1 point

Third possibility.

One solution is $x=b=6 k$.

The quadratic equation $2 x^{2}-x(a+3 b)+a^{2}+2 b^{2}-a b=0$ has two solutions and

$$
\begin{gathered}
x_{1}^{2}+x_{2}^{2}+x_{3}^{2}=4068 \\
\left(x_{2}-6\right)^{2}+x_{2}^{2}+\left(x_{2}+6\right)^{2}=4068 \\
3 x_{2}^{2}=4014
\end{gathered}
$$

This equation has no solutions in the set of natural numbers.

1 point

Therefore, the solution to the problem is

$$
a \cdot b=2772 \text { or } 1152
$$