Bài 3 trang 17 SGK Toán 8 Tập 1 Cánh diều - CD

a) Ta có: (x – y)(x² + xy + y²)

= x . x² + x . xy + x . y²– y . x² – y . xy– y . y²

= x³ + (x²y – x²y) + (xy²– xy²) – y³

= x³ – y³.

b) Ta có: (x + y)(x² – xy + y²)

= x . x² – x . xy + x . y² + y . x² – y . xy + y . y²

= x³ – x²y + xy² + x²y – xy² + y³

= x³ + (x²y – x²y) + (xy²– xy²) + y³

= x³ + y³.

c) Ta có: (4x - 1)(6y + 1) - 3x$\left(8x+\frac{4}{3}\right)$

= 4x.6y + 4x.1 - 1.6y - 1.1 - 3x.8x - 3x.$\frac{4}{3}$

= 24xy + 4x – 6y – 1 – 24x² – 4x

= 24xy – 24x² + (4x – 4x) – 6y – 1

= 24xy – 24x² – 6y – 1.

d) Ta có: (x + y)(x – y) + (xy⁴ – x³y²) : (xy²)

= x . x + x . y – x . y – y . y + (xy⁴) : (xy²) – (x³y²) : (xy²)

= x² – y² + y²– x²

= (x² – x²) + (y²– y²)

= 0.

-- Mod Toán 8 HỌC247