Find $k$ such that $(a+b)(b+c)(c+a)=(a+b+c)(a b+b c+c a)+k a b c$.

.
It is enough to develop the two members: $(a+b)(b+c)(c+a)=(a b+$ $\left.a c+b^{2}+b c\right)(c+a)=a^{2} b+a^{2} c+a b^{2}+a c^{2}+b^{2} c+b c^{2}+2 a b c,(a+b+c)(a b+b c+c a)+k a b c=$ $a^{2} b+a^{2} c+a b^{2}+a c^{2}+b^{2} c+b c^{2}+3 a b c+k a b c$ so $k=-1$.

## - Notable identities of degree 2 -

Proposition 1.3 (Notable identities of degree 2).

$$
\begin{aligned}
& (a+b)^{2}=a^{2}+2 a b+b^{2} \\
& (a-b)^{2}=a^{2}-2 a b+b^{2} \\
& a^{2}-b^{2}=(a+b)(a-b)
\end{aligned}
$$

## Proof.

![](https://cdn.mathpix.com/cropped/2024_05_10_59677aa68548da326d08g-129.jpg?height=650&width=1280&top_left_y=1862&top_left_x=342)
![](https://cdn.mathpix.com/cropped/2024_05_10_59677aa68548da326d08g-130.jpg?height=534&width=1264&top_left_y=263&top_left_x=404)