Đáp án:
$\begin{array}{l}
1)Do:{x^2} + {y^2} = 1\\
\Rightarrow {x^6} + 3{x^2}{y^2} + {y^6}\\
= {x^6} + {y^6} + 3{x^2}{y^2}\\
= {\left( {{x^2}} \right)^3} + {\left( {{y^2}} \right)^3} + 3{x^2}{y^2}\\
= \left( {{x^2} + {y^2}} \right)\left( {{x^4} - {x^2}{y^2} + {y^4}} \right) + 3{x^2}{y^2}\\
= 1.\left( {{x^4} - {x^2}{y^2} + {y^4}} \right) + 3{x^2}{y^2}\\
= {x^4} + 2{x^2}{y^2} + {y^4}\\
= {\left( {{x^2} + {y^2}} \right)^2}\\
= {1^2} = 1\\
Vay\,{x^6} + 3{x^2}{y^2} + {y^6} = 1\\
2)\\
{x^4} + {x^2}{y^2} + {y^4}\\
= {x^4} + 2{x^2}{y^2} + {y^4} - {x^2}{y^2}\\
= {\left( {{x^2} + {y^2}} \right)^2} - {x^2}{y^2}\\
= {\left( {{x^2} + {y^2}} \right)^2} - {\left( {xy} \right)^2}\\
= {a^2} - {b^2}\\
\left( {do:\left\{ \begin{array}{l}
{x^2} + {y^2} = a\\
xy = b
\end{array} \right.} \right)\\
3)\\
{\left( {{a^3} + {b^3} - {a^3}{b^3}} \right)^3} + 27{a^6}{b^6}\\
= {\left( {{a^3} + {b^3} - {{\left( {a + b} \right)}^3}} \right)^3} + 27{a^6}{b^6}\\
= {\left( {{a^3} + {b^3} - {a^3} - 3{a^2}b - 3a{b^2} - {b^3}} \right)^3} + 27{a^6}{b^6}\\
= {\left( { - 3{a^2}b - 3a{b^2}} \right)^3} + 27{a^6}{b^6}\\
= {\left[ { - 3ab\left( {a + b} \right)} \right]^3} + 27{a^6}{b^6}\\
= - 27{a^3}{b^3}.{\left( {a + b} \right)^3} + 27{a^6}{b^6}\\
= - 27{a^3}{b^3}.{\left( {ab} \right)^3} + 27{a^6}{b^6}\\
= - 27{a^6}{b^6} + 27{a^6}{b^6}\\
= 0\\
4)\\
a + b + c = 2p\\
\Rightarrow {p^2} + {\left( {p - a} \right)^2} + {\left( {p - b} \right)^2} + {\left( {p - c} \right)^2}\\
= \dfrac{1}{4}{\left( {a + b + c} \right)^2} + {\left( {\dfrac{1}{2}b + \dfrac{1}{2}c - \dfrac{1}{2}a} \right)^2}\\
+ {\left( {\dfrac{1}{2}a + \dfrac{1}{2}c - \dfrac{1}{2}b} \right)^2} + {\left( {\dfrac{1}{2}a + \dfrac{1}{2}b - \dfrac{1}{2}c} \right)^2}\\
= \dfrac{1}{4}\left( {{a^2} + {b^2} + {c^2} + 2ab + 2bc + 2ac} \right)\\
+ \dfrac{1}{4}\left( {{a^2} + {b^2} + {c^2} + 2bc - 2ac - 2ab} \right)\\
+ \dfrac{1}{4}\left( {{a^2} + {b^2} + {c^2} - 2ab - 2bc + 2ac} \right)\\
+ \dfrac{1}{4}\left( {{a^2} + {b^2} + {c^2} + 2ab - 2bc - 2ac} \right)\\
= {a^2} + {b^2} + {c^2}\\
Vay\,{p^2} + {\left( {p - a} \right)^2} + {\left( {p - b} \right)^2} + {\left( {p - c} \right)^2}\\
= {a^2} + {b^2} + {c^2}
\end{array}$
