Phân tích thành nhân tử 

- Hướng dẫn giải

$a\left(b^2+c^2\right)+b\left(c^2+a^2\right)+c\left(a^2+b^2\right)+2abc$

$$\begin{gathered} =a{b}^2+a{c}^2+b{c}^2+b{a}^2+c\left(a^2+b^2+2ab\right) \\ =\left(a{b}^2+a^{2}b\right)+\left(a{c}^2+b{c}^2\right)+c{\left(a+b\right)}^2 \\ =ab\left(a+b\right)+c^2\left(a+b\right)+c{\left(a+b\right)}^2 \\ =\left(a+b\right)\left(ab+c^2+ac+bc\right) \\ =\left(a+b\right)\left[\left(ab+ac\right)+\left(bc+c^2\right)\right] \\ =\left(a+b\right)\left[a\left(b+c\right)+c\left(b+c\right)\right] \\ =\left(a+b\right)\left(b+c\right)\left(c+a\right) \end{gathered}$$