Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
a,\\
{\left( {{x^2} - 3x - 1} \right)^2} - 12{x^2} + 36x + 39\\
 = {\left( {{x^2} - 3x - 1} \right)^2} + \left( { - 12{x^2} + 36x + 12} \right) + 27\\
 = {\left( {{x^2} - 3x - 1} \right)^2} - 12.\left( {{x^2} - 3x - 1} \right) + 27\\
 = \left[ {{{\left( {{x^2} - 3x - 1} \right)}^2} - 3.\left( {{x^2} - 3x - 1} \right)} \right] + \left[ { - 9\left( {{x^2} - 3x - 1} \right) + 27} \right]\\
 = \left( {{x^2} - 3x - 1} \right).\left[ {\left( {{x^2} - 3x - 1} \right) - 3} \right] - 9.\left[ {\left( {{x^2} - 3x - 1} \right) - 3} \right]\\
 = \left[ {\left( {{x^2} - 3x - 1} \right) - 3} \right].\left[ {\left( {{x^2} - 3x - 1} \right) - 9} \right]\\
 = \left( {{x^2} - 3x - 4} \right).\left( {{x^2} - 3x - 10} \right)\\
 = \left[ {\left( {{x^2} - 4x} \right) + \left( {x - 4} \right)} \right].\left[ {\left( {{x^2} - 5x} \right) + \left( {2x - 10} \right)} \right]\\
 = \left[ {x.\left( {x - 4} \right) + \left( {x - 4} \right)} \right].\left[ {x\left( {x - 5} \right) + 2.\left( {x - 5} \right)} \right]\\
 = \left( {x - 4} \right)\left( {x + 1} \right)\left( {x - 5} \right)\left( {x + 2} \right)\\
b,\\
{a^4} + {b^4} + {c^4} - 2{a^2}{b^2} - 2{b^2}{c^2} - 2{a^2}{c^2}\\
 = \left( {{a^4} - 2{a^2}{b^2} + {b^4}} \right) + \left( { - 2{b^2}{c^2} + 2{a^2}{c^2}} \right) + {c^4} - 4{a^2}{c^2}\\
 = {\left( {{a^2} - {b^2}} \right)^2} + 2{c^2}\left( {{a^2} - {b^2}} \right) + {c^4} - 4{a^2}{c^2}\\
 = \left[ {{{\left( {{a^2} - {b^2}} \right)}^2} + 2{c^2}\left( {{a^2} - {b^2}} \right) + {c^4}} \right] - 4{a^2}{c^2}\\
 = {\left( {{a^2} - {b^2} + {c^2}} \right)^2} - {\left( {2ac} \right)^2}\\
 = \left[ {\left( {{a^2} - {b^2} + {c^2}} \right) - 2ac} \right].\left[ {\left( {{a^2} - {b^2} + {c^2}} \right) + 2ac} \right]\\
 = \left[ {\left( {{a^2} - 2ac + {c^2}} \right) - {b^2}} \right].\left[ {\left( {{a^2} + 2ac + {c^2}} \right) - {b^2}} \right]\\
 = \left[ {{{\left( {a - c} \right)}^2} - {b^2}} \right].\left[ {{{\left( {a + c} \right)}^2} - {b^2}} \right]\\
 = \left( {a - c - b} \right)\left( {a - c + b} \right)\left( {a + c - b} \right)\left( {a + c + b} \right)
\end{array}\)