Question

Let  
P = a² − b² + 2ab,  
Q = a² + 4b² − 6ab,
R = b² + 6,  S = a² − 4ab and  
T = −2a² + b² − ab + a.
Find P + Q + R + S − T.

Answer 1

P = a² − b² + 2ab
Q = a² + 4b² − 6ab
R = b² + 6
S = a² − 4ab 
T = −2a² + b² − ab + a

Adding P, Q, R and S:
P+Q+R+S
= (a² − b² + 2ab)+(a² + 4b² − 6ab)+(b² + 6)+(a² − 4ab )
= a² − b² + 2ab+a² + 4b² − 6ab+b² + 6+a² − 4ab

Rearranging and collecting the like terms:
= (1+1+1)a² +(−1+4+1) b² + (2-6-4)ab+6
P+Q+R+S = 3a² +4b² - 8ab+6

To find P + Q + R + S − T, subtract T = (−2a² + b² − ab + a) from P+Q+R+S = (3a² +4b² - 8ab+6).

On changing the sign of each term of the expression that is to be subtracted and then adding:
Term to be subtracted = −2a² + b² − ab + a
Changing the sign of each term of the expression gives 2a² - b² + ab - a.
 Now add:
(3a² +4b² - 8ab+6)+(2a² - b² + ab - a) = 3a² +4b² - 8ab+6+2a² - b² + ab - a
= (3+2)a² +(4-1) b² +(-8+1) ab - a+6
 
P + Q + R + S − T = 5a² +3b² -7 ab - a+6