Question

if a = b and a'b'c then show that
b c (a+b+c) - c²

​

Answer 1

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