Câu hỏi của Hoàng Tuấn Linh

\(\left[x^3-3xy\left(x-y\right)-y^3\right]-x^2+2xy-y^2\)\(=\left(x^3-y^3\right)-3xy\left(x-y\right)-\left(x^2-2xy+y^2\right)\)\(=\left(x-y\right)\left(x^2+xy+y^2\right)-3xy\left(x-y\right)-\left(x-y\right)^2\)\(=\left(x-y\right)\left(x^2+xy+y^2-3xy-x+y\right)\)\(=\left(x-y\right)\left(x^2-2xy+y^2-x+y\right)\)\(=\left(x-y\right)\left[\left(x-y\right)^2-\left(x-y\right)\right]\)\(=\left(x-y\right)\left(x-y\right)\left(x-y-1\right)\)\(=\left(x-y\right)^2\left(x-y-1\right)\)\(=8^2\left(8-1\right)\\ =64\cdot7=448\)Câu trả lời: \(\left[x^3-3xy\left(x-y\right)-y^3\right]-x^2+2xy-y^2\)\(=\left(x^3-y^3\right)-3xy\left(x-y\right)-\left(x^2-2xy+y^2\right)\)\(=\left(x-y\right)\left(x^2+xy+y^2\right)-3xy\left(x-y\right)-\left(x-y\right)^2\)\(=\left(x-y\right)\left(x^2+xy+y^2-3xy-x+y\right)\)\(=\left(x-y\right)\left(x^2-2xy+y^2-x+y\right)\)\(=\left(x-y\right)\left[\left(x-y\right)^2-\left(x-y\right)\right]\)\(=\left(x-y\right)\left(x-y\right)\left(x-y-1\right)\)\(=\left(x-y\right)^2\left(x-y-1\right)\)\(=8^2\left(8-1\right)\\ =64\cdot7=448\)