Solve:

$(a+b)(a-b)+(b+c)(b-c)+(c+a)(c-a)$.

The given expression $(a+b)(a-b)+(b+c)(b-c)+(c+a)(c-a)$ can be solved as follows:

$\left[\right(a+b\left)\right(a-b\left)\right]+\left[\right(b+c\left)\right(b-c\left)\right]+\left[\right(c+a\left)\right(c-a\left)\right]=(a^2-b^2)+(b^2-c^2)+(c^2-a^2)(\because x^2-y^2=(x-y\left)\right(x+y\left)\right)=a^2-b^2+b^2-c^2+c^2-a^2=0$

Hence, $(a+b)(a-b)+(b+c)(b-c)+(c+a)(c-a)=0$.