Giải thích các bước giải:

$\begin{array}{l}
1){\left( {a + b + c} \right)^2} + {\left( {a + b - c} \right)^2} - 4{c^2}\\
 = {\left( {a + b + c} \right)^2} - {\left( {2c} \right)^2} + {\left( {a + b - c} \right)^2}\\
 = \left( {a + b - c} \right)\left( {a + b + 3c} \right) + {\left( {a + b - c} \right)^2}\\
 = \left( {a + b - c} \right)\left( {a + b + 3c + a + b - c} \right)\\
 = \left( {a + b - c} \right)\left( {2a + 2b + 2c} \right)\\
 = 2\left( {a + b - c} \right)\left( {a + b + c} \right)\\
2)4{a^2}{b^2} - {\left( {{a^2} + {b^2} - {c^2}} \right)^2}\\
 = {\left( {2ab} \right)^2} - {\left( {{a^2} + {b^2} - {c^2}} \right)^2}\\
 = \left( {2ab - \left( {{a^2} + {b^2} - {c^2}} \right)} \right)\left( {2ab + {a^2} + {b^2} - {c^2}} \right)\\
 = \left( {{c^2} - \left( {{a^2} + {b^2} - 2ab} \right)} \right)\left( {\left( {{a^2} + {b^2} + 2ab} \right) - {c^2}} \right)\\
 = \left( {{c^2} - {{\left( {a - b} \right)}^2}} \right)\left( {{{\left( {a + b} \right)}^2} - {c^2}} \right)\\
 = \left( {c - a + b} \right)\left( {c + a - b} \right)\left( {a + b - c} \right)\left( {a + b + c} \right)\\
 = \left( { - a + b + c} \right)\left( {a - b + c} \right)\left( {a + b - c} \right)\left( {a + b + c} \right)\\
3)a\left( {{b^2} - {c^2}} \right) + b\left( {{c^2} - {a^2}} \right) + c\left( {{a^2} - {b^2}} \right)\\
 = a\left( {b - c} \right)\left( {b + c} \right) + b{c^2} - {a^2}b + {a^2}c - {b^2}c\\
 = \left( {b - c} \right)\left( {ab + ac} \right) - bc\left( {b - c} \right) - {a^2}\left( {b - c} \right)\\
 = \left( {b - c} \right)\left( {ab + ac - bc - {a^2}} \right)\\
 = \left( {b - c} \right)\left( {b\left( {a - c} \right) - a\left( {a - c} \right)} \right)\\
 = \left( {b - c} \right)\left( {a - c} \right)\left( {b - a} \right)\\
4)ab\left( {a + b} \right) + bc\left( {b + c} \right) + ca\left( {c + a} \right) + 2abc\\
 = ab\left( {a + b} \right) + abc + bc\left( {b + c} \right) + abc + ca\left( {c + a} \right)\\
 = ab\left( {a + b + c} \right) + bc\left( {a + b + c} \right) + ca\left( {c + a} \right)\\
 = \left( {a + b + c} \right)\left( {ab + bc} \right) + ca\left( {c + a} \right)\\
 = \left( {a + b + c} \right)b\left( {a + c} \right) + ca\left( {c + a} \right)\\
 = \left( {a + c} \right)\left( {b\left( {a + b + c} \right) + ca} \right)\\
 = \left( {a + c} \right)\left( {ab + {b^2} + bc + ca} \right)\\
 = \left( {a + c} \right)\left( {a + b} \right)\left( {b + c} \right)\\
5)ab\left( {a + b} \right) + bc\left( {b + c} \right) + ca\left( {c + a} \right) + 3abc\\
 = \left( {ab\left( {a + b} \right) + abc} \right) + \left( {bc\left( {b + c} \right) + abc} \right) + \left( {ca\left( {c + a} \right) + abc} \right)\\
 = \left( {a + b + c} \right)\left( {ab + bc + ca} \right)
\end{array}$