if a+b+c=0, then prove that (b+c/bc)² + (c+a/ca)² + (a+b/ab)² = 0
Answers
Answer:
a² + b² + c² = ab + bc + ca (Given)
On multiplying both sides of the equation with 2, we get
2 ( a² + b² + c² ) = 2 ( ab + bc + ca)
2a² + 2b² + 2c² = 2ab + 2bc + 2ca
On expanding we get
=> a² + a² + b² + b² + c² + c² – 2ab – 2bc – 2ca = 0
Combining the like terms we get
=> a² + b² – 2ab + b² + c² – 2bc + c² + a² – 2ca = 0
From the identity we know that (a – b)2 = a2 – 2ab + b2 , we get
=> (a² + b² – 2ab) + (b² + c² – 2bc) + (c² + a² – 2ca) = 0
=>(a – b)² + (b – c)² + (c – a)² = 0
=>(a – b)² = (b – c)² = (c – a)² = 0
(a – b)² = 0 ———- (i)
(b – c)² = 0 ———- (ii)
(c – a)² = 0 ———- (iii)
On simplifying equation (i)we get
(a – b)² = 0
On taking Square Root on both sides, we get
a – b = 0
a = b ———- (iv)
On simplifying equation (ii) we get
(b – c)² = 0
On taking Square Root on both sides, we get
b – c = 0
b = c ———- (v)
On simplifying equation (iii) we get
(c – a)² = 0
On taking Square Root on both sides, we get
c – a = 0
c = a ———- (vi)
From Equation No. (iv), (v) & (vi) , we obtain
a = b = c