Respuesta :

devishri1977

Answer:

Step-by-step explanation:

ab² + 2a²b + a³ = a(b² + 2ab + a²)

Here variable 'a' is common in all the terms. HCF = a

2) 2ab² = 2 * a * b *b

   4a²b = 2 * 2 *a * a * b

6a³        = 2 * 3 * a * a * a

HCF = 2a

2ab² - 4a²b - 6a³ = 2a(b² - 2ab - 3a²)