Question

Prove that - -2a a+b c+a
a+b -2b b+c = 4(a-b)(b-c)(c-a)
c+a c+b -2c

Answer 1

Step-by-step explanation:

△=

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b+c

c+a

a+b

c+a

a+b

b+c

a+b

b+c

c+a

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⇒R

1

=R

1

+R

2

+R

3

⇒△=

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2(a+b+c)

a+c

a+b

2(a+b+c)

a+b

b+c

2(a+b+c)

b+c

c+a

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=2(a+b+c)

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1

c+a

a+b

1

a+b

b+c

1

b+c

c+a

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⇒C

2

=C

2

−C

1

;C

3

=C

3

−C

1

⇒△=2(a+b+c)

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1

c+a

a+b

0

b−c

c−a

0

b−a

c−b

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expanding along C

1

⇒△=2(a+b+c)×1

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b−c

c−a

b−a

c−b

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=2(a+b+c)[(b−c)(c−b)−(b−a)(c−a)]

=2(a+b+c)[bc−c

2

+bc−b

2

−bc+ac−a

2

+ab]

=2(a+b+c)(ab+bc+ca−a

2

−b

2

−c

2

)

Hence, proved.