Factoring $(a+b)(a+c)(b+c)=(a+b+c)(ab+bc+ca)-abc$ How to prove the following equality?

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

I did it
$$\begin{aligned}
a^2b + a^2c + ab^2 + cb^2 + bc^2 + ac^2 + 2abc &=a^2(b + c) + bc(b + c) + a(b^2 + c^2 +2bc)\\&=a^2(b + c) + bc(b+ c) +a(b + c)^2\\&=(b + c)(a^2 + bc + ab + ac)\\&=(b + c)\left[a(a + b) + c(a + b)\right]\\&=(a + b)(b + c)(c + a)
\end{aligned}$$
I just wanted to see some different way to work out with it.

Set $\displaystyle a+b+c=x$
So, we have $\displaystyle (x-a)(x-b)(x-c)=x^3-x^2(a+b+c)+(ab+bc+ca)x-abc$
Now, $\displaystyle  x^3-x^2(a+b+c)=x^2(x-\overline{a+b+c})=x^2\cdot0$
