1. Factorize the expression into prime factors

$$
T=(a b+b c+c a)(a+b+c)-a b c
$$
Solution. We have:

$$
\begin{aligned}
T & =a^{2} b+a^{2} c+a b c+a b^{2}+a b c+b^{2} c+a b c+a c^{2}+b c^{2}-a b c \\
& =\left(a^{2} b+a^{2} c\right)+\left(a b c+a b^{2}\right)+\left(b^{2} c+b c^{2}\right)+\left(a b c+a c^{2}\right) \\
& =a^{2}(b+c)+a b(b+c)+b c(b+c)+a c(b+c) \\
& =(b+c)\left(a^{2}+a b+b c+a c\right) \\
& =(b+c)[a(a+b)+c(a+b)] \\
& =(a+b)(b+c)(c+a)
\end{aligned}
$$