Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
1,\\
A = 8{x^3} - 12{x^2} + 6x - 1\\
 = {\left( {2x} \right)^3} - 3.{\left( {2x} \right)^2}.1 + 3.2x{.1^2} - {1^3}\\
 = {\left( {2x - 1} \right)^3}\\
x = \dfrac{1}{2} \Rightarrow A = {\left( {2.\dfrac{1}{2} - 1} \right)^3} = {0^3} = 0\\
2,\\
8{x^3} - {\left( {3x - 2} \right)^3}\\
 = {\left( {2x} \right)^3} - {\left( {3x - 2} \right)^3}\\
 = \left[ {2x - \left( {3x - 2} \right)} \right].\left[ {{{\left( {2x} \right)}^2} + 2x.\left( {3x - 2} \right) + {{\left( {3x - 2} \right)}^2}} \right]\\
 = \left( {2 - x} \right).\left( {4{x^2} + 6{x^2} - 4x + 9{x^2} - 12x + 4} \right)\\
 = \left( {2 - x} \right).\left( {19{x^2} - 16x + 4} \right)\\
3,\\
a,\\
62.53 + {31^2} + {53^2} - {16^2}\\
 = \left( {{{31}^2} + 62.53 + {{53}^2}} \right) - {16^2}\\
 = \left( {{{31}^2} + 2.31.53 + {{53}^2}} \right) - {16^2}\\
 = {\left( {31 + 53} \right)^2} - {16^2} = {84^2} - {16^2}\\
 = \left( {84 - 16} \right)\left( {84 + 16} \right)\\
 = 68.100 = 6800\\
b,\\
A = \left( {x - 4} \right)\left( {{x^2} + 4x + 16} \right) - x\left( {{x^2} + 3} \right)\\
 = \left( {x - 4} \right)\left( {{x^2} + x.4 + {4^2}} \right) - \left( {{x^3} + 3x} \right)\\
 = \left( {{x^3} - {4^3}} \right) - {x^3} - 3x\\
 =  - 3x - 64\\
x =  - \dfrac{2}{3} \Rightarrow A =  - 3.\left( { - \dfrac{2}{3}} \right) - 64 =  - 62\\
*)\\
\left( {x + 1} \right)\left( {{x^2} - x + 1} \right) - {\left( {x - 1} \right)^3} - 3x.\left( {x - 1} \right)\\
 = \left( {{x^3} + {1^3}} \right) - \left( {{x^3} - 3{x^2} + 3x - 1} \right) - \left( {3{x^2} - 3x} \right)\\
 = {x^3} + 1 - {x^3} + 3{x^2} - 3x + 1 - 3{x^2} + 3x\\
 = 2\\
*)\\
6{x^2} - 12x + 6 = 0\\
 \Leftrightarrow 6.\left( {{x^2} - 2x + 1} \right) = 0\\
 \Leftrightarrow 6.{\left( {x - 1} \right)^2} = 0\\
 \Leftrightarrow x - 1 = 0\\
 \Leftrightarrow x = 1\\
*)\\
{\left( {2x - 1} \right)^2} - {\left( {x + 3} \right)^2} = 0\\
 \Leftrightarrow \left[ {\left( {2x - 1} \right) - \left( {x + 3} \right)} \right].\left[ {\left( {2x - 1} \right) + \left( {x + 3} \right)} \right] = 0\\
 \Leftrightarrow \left( {x - 4} \right)\left( {3x + 2} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x - 4 = 0\\
3x + 2 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 4\\
x =  - \dfrac{2}{3}
\end{array} \right.
\end{array}\)