Question

from the sum of 2b²+3bc;-b²-bc-c² and bc+2c² subtract the sum of 3b²-c² and -b²+bc+c²​

Answer 1

Step-by-step explanation:

Given :-

2b²+3bc;

-b²-bc-c²

bc+2c²

and

3b²-c²

-b²+bc+c²

To find:-

From the sum of 2b²+3bc;-b²-bc-c² and bc+2c² subtract the sum of 3b²-c² and -b²+bc+c² ?

Solution :-

Given expressions are :

2b²+3bc;

-b²-bc-c²

bc+2c²

and

3b²-c²

-b²+bc+c²

Now,

On adding 2b²+3bc; -b²-bc-c² and bc+2c²

=> (2b²+3bc)+(-b²-bc-c²) +( bc+2c² )

=> (2b²-b²)+(3bc-bc+bc)+(-c²+2c²)

=> b²+3bc+c²

The sum of 2b²+3bc;-b²-bc-c² and bc+2c² is b²+3bc+c²

and

On adding 3b²-c² and -b²+bc+c²

=> (3b²-c²) + (-b²+bc+c²)

=> (3b²-b²)+(-c²+c²)+bc

=> 2b²+0+bc

=> 2b²+bc

The sum of 3b²-c² and -b²+bc+c² is 2b²+bc

On subtracting 2b²+bc from b²+3bc+c² then

=>( b²+3bc+c²)-(2b²+bc)

=> b²+3bc+c²-2b²-bc

=> (b²-2b²)+(3bc-bc)+c²

=> -b²+2bc+c²

Answer:-

The answer for the given problem is -b²+2bc+c²