1. Cho x-y=1. Tính x^3-y^3-3xy 2. Cho 3(a^2+b^2+c^2)=(a+b+c)^2 Cm: a=b=c 3. Cho a+b+c=abc , (1/a)+(1/b)+(1/c)=2 Tính (1/a^2)+(1/b^2)+(1/c^2)

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