Bài 1. Ta có: \(A = {\left( {2x} \right)^3} + {\left( {3y} \right)^3} = 8{x^3} + 27{y^3}.\)
Bài 2. Ta có:
\(\left( {4{x^2} + 2x + 1} \right)\left( {2x - 1} \right) - 4x\left( {2{x^2} - 3} \right)\)
\( = {\left( {2x} \right)^3} - {1^3} - 4x\left( {2{x^2} - 3} \right) \)
\(= 8{x^3} - 1 - 8{x^3} + 12x = 12x - 1\)
Vậy : \(12x - 1 = 23 \Rightarrow 12x = 24 \Rightarrow x = 2.\)
Bài 3. Ta có:
\({a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + ab + {b^2}} \right) \)
\(= \left( {a - b} \right)\left[ {{{\left( {a - b} \right)}^2} + 3ab} \right]\)
\( = 1.\left[ {{1^2} + 3.6} \right] = 19.\)
