Bài 1. Ta có:
\({\left( {{1 \over 2}a + b} \right)^3} + {\left( {{1 \over 2}a - b} \right)^3}\)
\(= {1 \over 8}{a^3} + {3 \over 4}{a^2}b + {3 \over 2}a{b^2} + {b^3} + {1 \over 8}{a^3} - {3 \over 4}{a^2}b + {3 \over 2}a{b^2}.\)
\(={1 \over 4}{a^3} + 3a{b^2}.\)
Bài 2. Ta có: \({x^3} - 3{x^2} + 3x - 1 = {\left( {x - 1} \right)^3}\)
Vậy: \({\left( {x - 1} \right)^3} = 0 \Rightarrow x - 1 = 0 \Rightarrow x = 1.\)
Bài 3. Ta có:
\({\left( {4x - 1} \right)^3} - \left( {4x - 3} \right)\left( {16{x^2} + 3} \right)\)
\( = \left( {64{x^3} - 48{x^2} + 12x - 1} \right)\)\(\; - \left( {64{x^3} + 12x - 48{x^2} - 9} \right)\)
\( = 64{x^3} - 48{x^2} + 12x - 1 - 64{x^3} \)\(\;- 12x + 48{x^2} + 9\)
\( = 8\) (không đổi).
