Đáp án:

\(\begin{array}{l}
1,\\
{\left( {a - b} \right)^3} + 3ab\left( {a - b} \right) = {a^3} - {b^3}\\
2,\\
 - 18{x^2} - 54\\
3,\\
{a^3} - {b^3} = 1 + 3ab\\
4,\\
x = 1\\
5,\\
8\\
6,\\
15{x^2} + 75x\\
7,\\
x =  - \dfrac{5}{{12}}\\
8,\\
 - 2\\
9,\\
x =  - 2\\
10,\\
{a^3} + {b^3} + {c^3} = 3abc\,\,\,\,\,\,\,\,\,\,\left( {a + b + c = 0} \right)\\
11,\\
{\left( {a + 2} \right)^3} - \left( {a + 6} \right)\left( {{a^2} + 12} \right) + 64 = 0\\
12,\\
A =  - 2{n^3}\\
13,\\
\left( {a - 1} \right)\left( {a - 2} \right)\left( {1 + a + {a^2}} \right)\left( {4 + 2a + {a^2}} \right) = {a^6} - 9{a^3} + 8\\
14,\\
x = 2
\end{array}\)

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
1,\\
{\left( {a - b} \right)^3} + 3ab\left( {a - b} \right)\\
 = \left( {{a^3} - 3{a^2}b + 3a{b^2} - {b^3}} \right) + \left( {3{a^2}b - 3a{b^2}} \right)\\
 = {a^3} + \left( { - 3{a^2}b + 3a{b^2}} \right) + \left( {3a{b^2} - 3a{b^2}} \right) - {b^3}\\
 = {a^3} + 0 + 0 - {b^3}\\
 = {a^3} - {b^3}\\
2,\\
{\left( {x - 3} \right)^3} - {\left( {x + 3} \right)^3}\\
 = \left( {{x^3} - 3.{x^2}.3 + 3.x{{.3}^2} - {3^3}} \right) - \left( {{x^3} + 3.{x^2}.3 + 3.x{{.3}^2} + {3^3}} \right)\\
 = \left( {{x^3} - 9{x^2} + 27x - 27} \right) - \left( {{x^3} + 9{x^2} + 27x + 27} \right)\\
 = {x^3} - 9{x^2} + 27x - 27 - {x^3} - 9{x^2} - 27x - 27\\
 =  - 18{x^2} - 54\\
3,\\
{a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + ab + {b^2}} \right)\\
 = 1.\left( {{a^2} + ab + {b^2}} \right)\\
 = {a^2} + ab + {b^2}\\
 = \left( {{a^2} - 2ab + {b^2}} \right) + 3ab\\
 = {\left( {a - b} \right)^2} + 3ab\\
 = {1^2} + 3ab\\
 = 1 + 3ab\\
4,\\
{x^3} - 3{x^2} + 3x - 1 = 0\\
 \Leftrightarrow {x^3} - 3.{x^2}.1 + 3.x{.1^2} - {1^3} = 0\\
 \Leftrightarrow {\left( {x - 1} \right)^3} = 0\\
 \Leftrightarrow x - 1 = 0\\
 \Leftrightarrow x = 1\\
5,\\
{\left( {4x - 1} \right)^3} - \left( {4x - 3} \right)\left( {16{x^2} + 3} \right)\\
 = \left[ {{{\left( {4x} \right)}^3} - 3.{{\left( {4x} \right)}^2}.1 + 3.4x{{.1}^2} - {1^3}} \right] - \left( {64{x^3} + 12x - 48{x^2} - 9} \right)\\
 = \left( {64{x^3} - 48{x^2} + 12x - 1} \right) - \left( {64{x^3} + 12x - 48{x^2} - 9} \right)\\
 = 64{x^3} - 48{x^2} + 12x - 1 - 64{x^3} - 12x + 48{x^2} + 9\\
 = 8\\
6,\\
{\left( {x + 5} \right)^3} - {x^3} - 125\\
 = \left( {{x^3} + 3.{x^2}.5 + 3.x{{.5}^2} + {5^3}} \right) - {x^3} - 125\\
 = \left( {{x^3} + 15{x^2} + 75x + 125} \right) - {x^3} - 125\\
 = 15{x^2} + 75x\\
7,\\
{\left( {x - 2} \right)^3} + 6.{\left( {x + 1} \right)^2} - {x^3} + 12 = 0\\
 \Leftrightarrow \left( {{x^3} - 3.{x^2}.2 + 3.x{{.2}^2} - {2^3}} \right) + 6.\left( {{x^2} + 2.x.1 + {1^2}} \right) - {x^3} + 12 = 0\\
 \Leftrightarrow \left( {{x^3} - 6{x^2} + 12x - 8} \right) + 6.\left( {{x^2} + 2x + 1} \right) - {x^3} + 12 = 0\\
 \Leftrightarrow {x^3} - 6{x^2} + 12x - 8 + 6{x^2} + 12x + 6 - {x^3} + 12 = 0\\
 \Leftrightarrow 24x + 10 = 0\\
 \Leftrightarrow 24x =  - 10\\
 \Leftrightarrow x =  - \dfrac{5}{{12}}\\
8,\\
{\left( {x - 1} \right)^3} - {x^3} + 3{x^2} - 3x - 1\\
 = \left( {{x^3} - 3.{x^2}.1 + 3.x{{.1}^2} - {1^3}} \right) - {x^3} + 3{x^2} - 3x - 1\\
 = \left( {{x^3} - 3{x^2} + 3x - 1} \right) - {x^3} + 3{x^2} - 3x - 1\\
 =  - 2\\
9,\\
{x^3} + 6{x^2} + 12x + 8 = 0\\
 \Leftrightarrow {x^3} + 3.{x^2}.2 + 3.x{.2^2} + {2^3} = 0\\
 \Leftrightarrow {\left( {x + 2} \right)^3} = 0\\
 \Leftrightarrow x + 2 = 0\\
 \Leftrightarrow x =  - 2\\
10,\\
{a^3} + {b^3} + {c^3}\\
 = \left( {{a^3} + 3{a^2}b + 3a{b^2} + {b^3}} \right) - 3{a^2}b - 3a{b^2} + {c^3}\\
 = {\left( {a + b} \right)^3} - \left( {3{a^2}b + 3a{b^2}} \right) + {c^3}\\
 = \left[ {{{\left( {a + b} \right)}^3} + {c^3}} \right] - \left( {3{a^2}b + 3a{b^2}} \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} \right)\\
 = \left( {a + b + c} \right).\left[ {{{\left( {a + b} \right)}^2} - \left( {a + b} \right).c + {c^2}} \right] - 3ab\left( {a + b} \right)\\
 = 0.\left[ {{{\left( {a + b} \right)}^2} - \left( {a + b} \right).c + {c^2}} \right] - 3ab\left( {a + b} \right)\\
 =  - 3ab\left( {a + b} \right)\\
a + b + c = 0 \Rightarrow a + b =  - c\\
 \Rightarrow  - 3ab\left( {a + b} \right) =  - 3ab.\left( { - c} \right) = 3abc\\
 \Rightarrow {a^3} + {b^3} + {c^3} = 3abc\\
11,\\
{\left( {a + 2} \right)^3} - \left( {a + 6} \right)\left( {{a^2} + 12} \right) + 64\\
 = \left( {{a^3} + 3.{a^2}.2 + 3.a{{.2}^2} + {2^3}} \right) - \left( {{a^3} + 12a + 6{a^2} + 72} \right) + 64\\
 = \left( {{a^3} + 6{a^2} + 12a + 8} \right) - \left( {{a^3} + 12a + 6{a^2} + 72} \right) + 64\\
 = {a^3} + 6{a^2} + 12a + 8 - {a^3} - 12a - 6{a^2} - 72 + 64\\
 = 0\\
12,\\
A = \left( {m - n} \right)\left( {{m^2} + mn + {n^2}} \right) - \left( {m + n} \right).\left( {{m^2} - mn + {n^2}} \right)\\
 = \left( {{m^3} - {n^3}} \right) - \left( {{m^3} + {n^3}} \right)\\
 = {m^3} - {n^3} - {m^3} - {n^3}\\
 =  - 2{n^3}\\
13,\\
\left( {a - 1} \right)\left( {a - 2} \right)\left( {1 + a + {a^2}} \right)\left( {4 + 2a + {a^2}} \right)\\
 = \left[ {\left( {a - 1} \right)\left( {1 + a + {a^2}} \right)} \right].\left[ {\left( {a - 2} \right)\left( {4 + 2a + {a^2}} \right)} \right]\\
 = \left[ {\left( {a - 1} \right).\left( {{a^2} + a.1 + {1^2}} \right)} \right].\left[ {\left( {a - 2} \right).\left( {{a^2} + a.2 + {2^2}} \right)} \right]\\
 = \left( {{a^3} - {1^3}} \right).\left( {{a^3} - {2^3}} \right)\\
 = \left( {{a^3} - 1} \right)\left( {{a^3} - 8} \right)\\
 = {a^6} - 8{a^3} - {a^3} + 8\\
 = {a^6} - 9{a^3} + 8\\
14,\\
\left( {x + 2} \right)\left( {{x^2} - 2x + 4} \right) - x\left( {x - 3} \right)\left( {x + 3} \right) = 26\\
 \Leftrightarrow \left( {x + 2} \right).\left( {{x^2} - x.2 + {2^2}} \right) - x.\left( {{x^2} - {3^2}} \right) = 26\\
 \Leftrightarrow \left( {{x^3} + {2^3}} \right) - \left( {{x^3} - 9x} \right) = 26\\
 \Leftrightarrow {x^3} + 8 - {x^3} + 9x = 26\\
 \Leftrightarrow 9x + 8 = 26\\
 \Leftrightarrow 9x = 18\\
 \Leftrightarrow x = 2
\end{array}\)