\[\begin{array}{l}
1)\,\,\,x - y = 1\\
{x^3} - {y^3} - 3xy = \left( {x - y} \right)\left( {{x^2} + xy + {y^2}} \right) - 3xy\\
= {x^2} + xy + {y^2} - 3xy = {x^2} - 2xy + {y^2} = {\left( {x - y} \right)^2} = 1.\\
2)\,\,\,3\left( {{a^2} + {b^2} + {c^2}} \right) = {\left( {a + b + c} \right)^2}\\
\Leftrightarrow 3{a^2} + 3{b^2} + 3{c^2} = {a^2} + 2ab + {b^2} + 2bc + {c^2} + 2ac\\
\Leftrightarrow 2{a^2} - 2ab + 2{b^2} - 2bc + 2{c^2} - 2ac = 0\\
\Leftrightarrow {a^2} - 2ab + {b^2} + {b^2} - 2bc + {c^2} + {c^2} - 2ac + {c^2} = 0\\
\Leftrightarrow {\left( {a - b} \right)^2} + {\left( {b - c} \right)^2} + {\left( {c - a} \right)^2} = 0\\
\Leftrightarrow \left\{ \begin{array}{l}
a - b = 0\\
b - c = 0\\
c - a = 0
\end{array} \right. \Leftrightarrow a = b = c.\\
3)\,\,\,a + b + c = abc;\,\,\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 2\\
Ta\,\,\,co:\\
{\left( {\,\frac{1}{a} + \frac{1}{b} + \frac{1}{c}} \right)^2} = 4\\
\Leftrightarrow \frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} + \frac{1}{{{c^2}}} + \frac{2}{{ab}} + \frac{2}{{bc}} + \frac{2}{{ca}} = 4\\
\Leftrightarrow \frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} + \frac{1}{{{c^2}}} + 2\left( {\frac{c}{{abc}} + \frac{a}{{abc}} + \frac{b}{{abc}}} \right) = 4\\
\Leftrightarrow \frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} + \frac{1}{{{c^2}}} + \frac{{2\left( {a + b + c} \right)}}{{abc}} = 4\\
\Leftrightarrow \frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} + \frac{1}{{{c^2}}} + 2 = 4\\
\Leftrightarrow \frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} + \frac{1}{{{c^2}}} = 2.
\end{array}\]
