if a = b and a'b'c then show that
b c (a+b+c) - c²
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Answer:
Consider a
2
+b
2
+c
2
−ab−bc−ca=0
Multiply both sides with 2, we get
2(a
2
+b
2
+c
2
−ab−bc−ca)=0
⇒2a
2
+2b
2
+2c
2
−2ab−2bc−2ca=0
⇒a
2
+a
2
+b
2
+b
2
+c
2
+c
2
−2ab−2bc−2ca=0
⇒(a
2
−2ab+b
2
)+(b
2
−2bc+c
2
)+(c
2
−2ca+a
2
)=0
⇒(a−b)
2
+(b−c)
2
+(c−a)
2
=0
Since the sum of squares is zero, it means 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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