a, x ² – 16 – 4xy + 4y ² b, x ² – 3x ³ – x +3 c, x ³ – x ²y – xy ² + y ³ d, 3x + 3y – x ² – 2xy – y ² e, 4x^4 + 4x ³ – x ² – x f, x^4 – 4x ³ + 8x ² –

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1. Giải thích các bước giải:

   Ta có:

   \(\begin{array}{l}
   a,\\
   {x^2} – 16 – 4xy + 4{y^2}\\
    = \left( {{x^2} – 4xy + 4{y^2}} \right) – 16\\
    = {\left( {x – 2y} \right)^2} – {4^2}\\
    = \left( {x – 2y – 4} \right)\left( {x – 2y + 4} \right)\\
   b,\\
   {x^2} – 3{x^3} – x + 3\\
    = \left( { – 3{x^3} + 3} \right) + \left( {{x^2} – x} \right)\\
    =  – 3.\left( {{x^3} – 1} \right) + x\left( {x – 1} \right)\\
    =  – 3.\left( {x – 1} \right)\left( {{x^2} + x + 1} \right) + x\left( {x – 1} \right)\\
    = \left( {x – 1} \right).\left[ { – 3\left( {{x^2} + x + 1} \right) + x} \right]\\
    = \left( {x – 1} \right)\left( { – 3{x^2} – 2x – 3} \right)\\
   c,\\
   {x^3} – {x^2}y – x{y^2} + {y^3}\\
    = \left( {{x^3} – {x^2}y} \right) – \left( {x{y^2} – {y^3}} \right)\\
    = {x^2}\left( {x – y} \right) – {y^2}\left( {x – y} \right)\\
    = \left( {x – y} \right)\left( {{x^2} – {y^2}} \right)\\
    = \left( {x – y} \right)\left( {x – y} \right)\left( {x + y} \right)\\
    = {\left( {x – y} \right)^2}\left( {x + y} \right)\\
   d,\\
   3x + 3y – {x^2} – 2xy – {y^2}\\
    = \left( {3x + 3y} \right) – \left( {{x^2} + 2xy + {y^2}} \right)\\
    = 3.\left( {x + y} \right) – {\left( {x + y} \right)^2}\\
    = \left( {x + y} \right)\left( {3 – x – y} \right)\\
   e,\\
   4{x^4} + 4{x^3} – {x^2} – x\\
    = \left( {4{x^4} + 4{x^3}} \right) – \left( {{x^2} + x} \right)\\
    = 4{x^3}\left( {x + 1} \right) – x\left( {x + 1} \right)\\
    = \left( {x + 1} \right)\left( {4{x^3} – x} \right)\\
    = x.\left( {x + 1} \right)\left( {4{x^2} – 1} \right)\\
    = x\left( {x + 1} \right)\left( {2x – 1} \right)\left( {2x + 1} \right)\\
   f,\\
   {x^4} – 4{x^3} + 8{x^2} – 16x + 16\\
    = \left( {{x^4} – 4{x^3} + 4{x^2}} \right) + \left( {4{x^2} – 16x + 16} \right)\\
    = {x^2}\left( {{x^2} – 4x + 4} \right) + 4\left( {{x^2} – 4x + 4} \right)\\
    = \left( {{x^2} – 4x + 4} \right)\left( {{x^2} + 4} \right)\\
    = {\left( {x – 2} \right)^2}.\left( {{x^2} + 4} \right)\\
   g,\\
   {a^3} + {b^3} + {c^3} – 3abc\\
    = \left( {{a^3} + 3{a^2}b + 3a{b^2} + {b^3}} \right) + {c^3} – \left( {3{a^2}b + 3a{b^2} + 3abc} \right)\\
    = {\left( {a + b} \right)^3} + {c^3} – 3ab\left( {a + b + c} \right)\\
    = \left[ {\left( {a + b} \right) + c} \right].\left[ {{{\left( {a + b} \right)}^2} – \left( {a + b} \right).c + {c^2}} \right] – 3ab\left( {a + b + c} \right)\\
    = \left( {a + b + c} \right)\left( {{a^2} + 2ab + {b^2} – ac – bc + {c^2}} \right) – 3ab\left( {a + b + c} \right)\\
    = \left( {a + b + c} \right)\left( {{a^2} + 2ab + {b^2} – ac – bc + {c^2} – 3ab} \right)\\
    = \left( {a + b + c} \right)\left( {{a^2} + {b^2} + {c^2} – ab – bc – ca} \right)
   \end{array}\)

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