What are the integers $k$ such that for all real numbers $a, b, c$,

$$
(a+b+c)(a b+b c+c a)+k a b c=(a+b)(b+c)(c+a)
$$
Suppose the equation is verified for an integer $k$. By taking $a=b=c=1$, we get $3 \times 3 + k = 2^{3}$, so $k = 8 - 9 = -1$.

It remains to verify that $(a+b+c)(ab+bc+ca) - abc = (a+b)(b+c)(c+a)$ for all real numbers $a, b, c$. Now,

$$
(a+b+c)(ab+bc+ca) - abc = a^2b + a^2c + abc + b^2c + b^2a + abc + c^2a + c^2b + abc - abc
$$

so

$$
(a+b+c)(ab+bc+ca) - abc = a^2b + a^2c + b^2a + b^2c + c^2a + c^2b + 2abc
$$

and

$$
(a+b)(b+c)(a+c) = (ab + b^2 + ac + bc)(a+c) = a^2b + ab^2 + a^2c + abc + abc + b^2c + ac^2 + bc^2
$$

so

$$
(a+b)(b+c)(a+c) = a^2b + a^2c + b^2a + b^2c + c^2a + c^2b + 2abc
$$

For $k = -1$, we indeed have for all real numbers $a, b, c$,

$$
(a+b+c)(ab+bc+ca) + kabc = (a+b)(b+c)(c+a)
$$

The unique solution is therefore $k = -1$.