Giải thích các bước giải:

3,

\(\begin{array}{l}
{\left( {a - b} \right)^2} + {\left( {b - c} \right)^2} + {\left( {c - a} \right)^2} = {\left( {a + b - 2c} \right)^2} + {\left( {b + c - 2a} \right)^2} + {\left( {c + a - 2b} \right)^2}\\
 \Leftrightarrow \left[ {{{\left( {a - b} \right)}^2} - {{\left( {a + b - 2c} \right)}^2}} \right] + \left[ {{{\left( {b - c} \right)}^2} - {{\left( {b + c - 2a} \right)}^2}} \right] + \left[ {{{\left( {c - a} \right)}^2} - {{\left( {c + a - 2b} \right)}^2}} \right] = 0\\
 \Leftrightarrow \left[ {\left( {a - b} \right) - \left( {a + b - 2c} \right)} \right].\left[ {\left( {a - b} \right) + \left( {a + b - 2c} \right)} \right] + \left[ {\left( {b - c} \right) - \left( {b + c - 2a} \right)} \right].\left[ {\left( {b - c} \right) + \left( {b + c - 2a} \right)} \right] + \left[ {\left( {c - a} \right) - \left( {c + a - 2b} \right)} \right].\left[ {\left( {c - a} \right) + \left( {c + a - 2b} \right)} \right] = 0\\
 \Leftrightarrow \left( {2c - 2b} \right)\left( {2a - 2c} \right) + \left( {2a - 2c} \right).\left( {2b - 2a} \right) + \left( {2b - 2a} \right).\left( {2c - 2b} \right) = 0\\
 \Leftrightarrow \left( {c - b} \right)\left( {a - c} \right) + \left( {a - c} \right)\left( {b - a} \right) + \left( {b - a} \right).\left( {c - b} \right) = 0\\
 \Leftrightarrow \left( {a - c} \right).\left[ {\left( {c - b} \right) + \left( {b - a} \right)} \right] + \left( {b - a} \right)\left( {c - b} \right) = 0\\
 \Leftrightarrow \left( {a - c} \right).\left( {c - a} \right) + \left( {b - a} \right)\left( {c - b} \right) = 0\\
 \Leftrightarrow  - {a^2} - {c^2} + 2ac + bc - {b^2} - ac + ab = 0\\
 \Leftrightarrow  - {a^2} - {c^2} + ac + bc - {b^2} + ab = 0\\
 \Leftrightarrow {a^2} + {b^2} + {c^2} - ab - bc - ca = 0\\
 \Leftrightarrow 2{a^2} + 2{b^2} + 2{c^2} - 2ab - 2bc - 2ca = 0\\
 \Leftrightarrow \left( {{a^2} - 2ab + {b^2}} \right) + \left( {{b^2} - 2bc + {c^2}} \right) + \left( {{c^2} - 2ca + {a^2}} \right) = 0\\
 \Leftrightarrow {\left( {a - b} \right)^2} + {\left( {b - c} \right)^2} + {\left( {c - a} \right)^2} = 0\\
 \Rightarrow \left\{ \begin{array}{l}
{\left( {a - b} \right)^2} = 0\\
{\left( {b - c} \right)^2} = 0\\
{\left( {c - a} \right)^2} = 0
\end{array} \right. \Leftrightarrow a = b = c\\
4,\\
\left( {a + b + c + d} \right)\left( {a - b - c + d} \right) = \left( {a - b + c - d} \right).\left( {a + b - c - d} \right)\\
 \Leftrightarrow \left[ {\left( {a + d} \right) + \left( {b + c} \right)} \right].\left[ {\left( {a + d} \right) - \left( {b + c} \right)} \right] = \left[ {\left( {a - d} \right) - \left( {b - c} \right)} \right].\left[ {\left( {a - d} \right) + \left( {b - c} \right)} \right]\\
 \Leftrightarrow {\left( {a + d} \right)^2} - {\left( {b + c} \right)^2} = {\left( {a - d} \right)^2} - {\left( {b - c} \right)^2}\\
 \Leftrightarrow \left( {{a^2} + 2ad + {d^2}} \right) - \left( {{b^2} + 2bc + {c^2}} \right) = \left( {{a^2} - 2ad + {d^2}} \right) - \left( {{b^2} - 2bc + {c^2}} \right)\\
 \Leftrightarrow 2ad - 2bc =  - 2ad + 2bc\\
 \Leftrightarrow 4ad = 4bc\\
 \Leftrightarrow ad = bc\\
 \Leftrightarrow \frac{a}{c} = \frac{b}{d}
\end{array}\)