$\begin{array}{l} {x^6} + {\left( {{x^3} - 3} \right)^3} = 3{x^5} - 9{x^2} - 1\\ \Leftrightarrow {x^6} + {\left( {{x^3} - 3} \right)^3} - 3{x^5} + 9{x^2} + 1 = 0\\ \Leftrightarrow {\left( {{x^2}} \right)^3} + {\left( {{x^3} - 3} \right)^3} - 3{x^2}\left( {{x^3} - 3} \right) + 1 = 0\\ a = {x^2},\,\,b = {x^3} - 3\\ PT \Leftrightarrow {a^3} + {b^3} - 3ab + 1 = 0\\ \Leftrightarrow {\left( {a + b} \right)^3} - 3ab\left( {a + b} \right) - 3ab + 1 = 0\\ \Leftrightarrow {\left( {a + b + 1} \right)^3} - 3ab\left( {a + b + 1} \right) - 3\left( {a + b} \right)\left( {a + b + 1} \right) = 0\\ \Leftrightarrow \left( {a + b + 1} \right)\left[ {{{\left( {a + b + 1} \right)}^2} - 3ab - 3\left( {a + b} \right)} \right] = 0\\ \Leftrightarrow \left( {a + b + 1} \right)\left( {{a^2} + {b^2} + 2ab + 2a + 2b - 3ab - 3a - 3b} \right) = 0\\ \Leftrightarrow \dfrac{1}{2}\left( {a + b + 1} \right)\left[ {{{\left( {a - 1} \right)}^2} + {{\left( {b - 1} \right)}^2} + {{\left( {a - b} \right)}^2}} \right] = 0\\ \Leftrightarrow \left[ \begin{array}{l} a + b + 1 = 0\\ a = b = 1 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} {x^2} + {x^3} - 3 + 1 = 0\\ \left\{ \begin{array}{l} {x^2} = 1\\ {x^3} = 4 \end{array} \right.(L) \end{array} \right.\\ \Rightarrow {x^3} + {x^2} - 2 = 0 \Leftrightarrow {x^3} - {x^2} + 2{x^2} - 2 = 0\\ \Leftrightarrow {x^2}\left( {x - 1} \right) + 2\left( {{x^2} - 1} \right) = 0\\ \Leftrightarrow \left( {x - 1} \right)\left( {{x^2} + 2x + 2} \right) = 0\\ \Leftrightarrow x = 1\\ \Rightarrow S = \left\{ 1 \right\} \end{array}$
