if 3 - 2 b + C is equal to zero then the value of 9 a square minus b square + c square + 6 i.
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2p
2
q
2
−3pq+4,5+7pq−3p
2
q
2
Adding we get,
2p
2
q
2
−3pq+4+5+7pq−3p
2
q
2
=−p
2
q
2
+4pq+9
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Answer:
Consider, a2+b2+c2–ab–bc–ca=0
Multiply both sides with 2, we get
2(a2+b2+c2–ab–bc–ca)=0
⇒ 2a2+2b2+2c2–2ab–2bc–2ca=0
⇒ (a2–2ab+b2)+(b2–2bc+c2)+(c2–2ca+a2)=0
⇒ (a–b)2+(b–c)2+(c–a)2=0
Since the sum of square is zero then each term should be zero
⇒ (a–b)2=0,(b–c)2=0,(c–a)2=0
⇒ (a–b)=0,(b–c)=0,(c–a)=0
⇒ a=b,b=c,c=a
∴ a=b=c.
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