Đáp án:
$\begin{array}{l}
A = \left( {x + 5} \right)\left( {{x^2} - 5x + 25} \right)\\
- {\left( {2x + 1} \right)^3} + 7{\left( {x - 1} \right)^3} - 3x\left( { - 11x + 5} \right)\\
= {x^3} + {5^3} - \left( {8{x^3} + 12{x^2} + 6x + 1} \right)\\
+ 7\left( {{x^3} - 3{x^2} + 3x - 1} \right) + 33{x^2} - 15x\\
= {x^3} + 125 - 8{x^3} - 12{x^2} - 6x - 1\\
+ 7{x^3} - 21{x^2} + 21x - 7 + 33{x^2} - 15x\\
= 117\\
B = {\left( {3x + 2} \right)^3} - 18x\left( {3x + 2} \right) + {\left( {x - 1} \right)^3}\\
- 28{x^3} + 3x\left( {x - 1} \right)\\
= 27{x^3} + 54{x^2} + 36x + 8\\
- 54{x^2} - 36x + {x^3} - 3{x^2} + 3x - 1 - 28{x^3}\\
+ 3{x^2} - 3x\\
= 7\\
C = \left( {4x - 1} \right)\left( {16{x^2} + 4x + 1} \right) - {\left( {4x + 1} \right)^3}\\
+ 12\left( {4{x^2} + 1} \right) - 15\\
= {\left( {4x} \right)^3} - 1 - 64{x^3} - 48{x^2} - 12x - 1\\
+ 48{x^2} + 12 - 15\\
= - 5
\end{array}$
