Question

answer me this both tq for all answering person​

Answer 1

Question

• (a² - b²) + (b² - c²) + (c² - a²)
• a³ - 2ab - (2a³ + 4b) - (9ab - 3b)

Solution

i) (a² - b²) + (b² - c²) + (c² - a²)

= a² - b² + b² - c² + c² - a²

= a² - a² - b² + b² - c² + c²

= 0

ii) a³ - 2ab - (2a³ + 4b) - (9ab - 3b)

= a³ - 2ab - 2a³ - 4b - 9ab + 3b

= a³ - 2a³ - 2ab - 9ab - 4b + 3b

= -a³ - 11ab - b

Take (-) minus as a common

= - (a³ + 11ab + b)

Note

• Same sign = addition
• Opposite sign = substraction

Answer 2

Solution

1) (a² - b²) + (b² - c²) + (c² - a²)

= a² - b² + b² - c² + c² - a²

= a² - a² - b² + b² - c² + c²

= 0

_____________________

2) a³ - 2ab - (2a³ + 4b) - (9ab - 3b)

= a³ - 2ab - 2a³ + 4b - 9ab + 3b

= a³ - 2a³ - 2ab - 9ab + 4b + 3b

= - a³ - 11ab + 7b