Question

if a+b+c=0 then proof that (b+c)²\bc + (c+a)²/ca +(a+b)²/ab = 3​

Answer 1

Step-by-step explanation:

Given :-

a + b + c = 0

To find :-

Prove that :

(b+c)²/bc + (c+a)²/ca +(a+b)²/ab = 3

Solution :-

Given that :-

a + b + c = 0 ----------(1)

=> b + c = - a ---------(2)

or c + a = - b --------(3)

or a + b = - c ---------(4)

We know that

If a+b+c=0 then a³+b³+c³=3abc ----------(5)

Now,

On taking LHS :-

(b+c)²/bc + (c+a)²/ca +(a+b)²/ab

From (2),(3)&(4)

=> [(-a)²/bc] + [ (-b)²/ca ] + [ (-c)²/ab]

=> (a²/bc)+(b²/ca)+(c²/ab)

LCM of bc , ac, ab is abc

=> [(a²×a)+(b²×b)+(c²×c)]/(abc)

=> (a³+b³+c³)/(abc)

Since a^m×a^n = a^(m+n)

From (5)

=> (3abc)/(abc)

=> 3

=> RHS

=> LHS = RHS

Hence, Proved.

Answer:-

If a + b + c = 0 then

[(b+c)²/bc] + [(c+a)²/ca] +[(a+b)²/ab] = 3

Used formulae:-

• If a + b + c = 0 then a³+b³+c³ = 3abc

• a^m×a^n = a^(m+n)