Bài 1. Ta có:
\(A = x\left( {{x^2} - 4} \right) - \left( {{x^3} - 27} \right)\)
\(\;\;\;\;= {x^3} - 4x - {x^3} + 27 = - 4x + 27\)
Với \(x = {1 \over 4},\) ta có: \(A = \left( { - 4} \right).{1 \over 4} + 27 = 26.\)
Bài 2. Ta có:
\(\left( {4x + 1} \right)\left( {16{x^2} - 4x + 1} \right) - 16x\left( {4{x^2} - 5} \right)\)
\( = \left( {64{x^3} + {1^3}} \right) - 64{x^3} + 80x \)
\(= 80x + 1\)
Vậy \(80x + 1 = 17 \Rightarrow x = {1 \over 5}.\)
Bài 3. Ta có:
\(P = \left( {a + 1} \right)\left( {{a^2} - a + 1} \right)\left( {a + 1} \right)\left( {{a^2} + a + 1} \right) \)
\(\;\;\;\;= \left( {{a^3} + 1} \right)\left( {{a^3} - 1} \right) = {a^6} - 1.\)
