Bài 1.
a) \(\left( {64{a^3} + 125{b^3}} \right) + 5b\left( {16{a^2} - 25{b^2}} \right) \)
\(= {\left( {4a} \right)^3} + {\left( {5b} \right)^3} + 5b\left[ {{{\left( {4a} \right)}^2} - {{\left( {5b} \right)}^2}} \right]\)
\( = \left( {4a + 5b} \right)\left( {16{a^2} - 20ab + 25{b^2}} \right) + 5b\left( {4a + 5b} \right)\left( {4a - 5b} \right)\)
\( = \left( {4a + 5b} \right)\left( {16{a^2} - 20ab + 25{b^2} + 20ab - 25{b^2}} \right)\)
\(= 16{a^2}\left( {4a - 5b} \right)\)
b) \(1 - \left( {{x^2} - 2xy + {y^2}} \right) = {1^2} - {\left( {x - y} \right)^2} \)
\(= \left( {1 + x - y} \right)\left( {1 - x + y} \right).\)
c) \({x^6} - 1 = \left( {{x^3} - 1} \right)\left( {{x^3} + 1} \right) \)
\(= \left( {x - 1} \right)\left( {{x^2} + x + 1} \right)\left( {x + 1} \right)\left( {{x^2} - x + 1} \right).\)
Bài 2. Ta có:
\(4{x^3} - 36x = 4x\left( {{x^2} - 9} \right) \)
\(= 4x\left( {x - 3} \right)\left( {x + 3} \right)\)
Vậy \(x\left( {x - 3} \right)\left( {x + 3} \right) = 0\)
\(\Rightarrow x = 0;x - 3 = 0;x+3=0\)
\(\Rightarrow x=0; x= 3; x=-3\)
Bài 3. Ta có:
\({\left( {4n - 3} \right)^2} - {\left( {3n - 4} \right)^2} \)
\(= \left( {4n - 3 + 3n - 4} \right)\left( {4n - 3 - 3n + 4} \right)\)
\(= \left( {7n - 7} \right)\left( {n + 1} \right) \)
\(= 7\left( {n - 1} \right)\left( {n + 1} \right) \;\vdots\; 7\) (với mọi giá trị n thuộc \(\mathbb Z\) ).
