Question

subtract the sum of ab²-3c²and 2b²+bc-2c² from the sum of 2b²2bc-c²and c+²b(b)​

Answer 1

Sum of ab² - 3c² + 2b² + bc - 2c²

= ab² (-3c² -2c²) + 2b² - bc

= ab² - 5c² + 2b² - bc

Sum of 2b² + 2bc - c² + c² + b²

= (2b² + b²) + 2bc (-c² + c²)

= 3b² + 2bc

Subtract ab² - 5c² + 2b² - bc from 3b² + 2bc

= 3b² + 2bc - ab² - 5c² + 2b² - bc

= (3b² + 2b²) + (2bc - bc) - ab² - 5c²

= 5b² + 1bc - ab² - 5c²

Hope it helps.

Answer 2

Step-by-step explanation:

ANSWER

(6a+4b−c+3)+(2b−3c+4)+(11b−7a+2c−1)+(2c−5a−6)

⇒ 6a+4b−c+3+2b−3c+4+11b−7a+2c−1+2c−5a−6

Now arranging and combining the like terms,

⇒ 6a−7a−5a+4b+2b+11b−c−3c+2c+2c+3+4−1−6

⇒ −6a+17b

∴ The sum of (6a+4b−c+3),(2b−3c+4),(11b−7a+2c−1) and (2c−5a−6) is (−6a+17b)