Question

(b+c)²/bc + (c+a)²/ca+(a+b)²/ab=3

Answer 1

Correct Question:

If a+b+c =0 then show $\sf\Large\ \frac{(b+c)^2}{bc} + \frac{(c+a)^2}{ca} + \frac{(a+b)^2}{ab}$=3.

Solution:

Given:

= a + b + c=0

• a + b = -c
• a + c = -a
• c + a = -1

To Prove:

$\sf\Large\ \frac{(b+c)^2}{bc} + \frac{(c+a)^2}{ca} + \frac{(a+b)^2}{ab}$

Solution:

Taking LHS,

$\sf\Large\ \frac{(b+c)^2}{bc} + \frac{(c+a)^2}{ca} + \frac{(a+b)^2}{ab}$=3

Putting the value of eq. 1,2,3

➡$\sf\Large\ \frac{(-a)^2}{bc} + \frac{(-b)^2}{ca} + \frac{(-c)^2}{ab}$

➡$\sf\Large\ \frac{a^2}{bc} + \frac{b^2}{ca} + \frac{c^2}{ab}$

Now we taking LCM,

➡$\sf\Large\ \frac{a^3+b^3+c^3}{abc}$

We know that if

➡$\sf\ (a+b+c)=0 , a^3+b^3+c^3=3abc$

So,

➡$\sf\Large\ \frac{3abc}{abc}$

$\dashrightarrow\: \underline{\boxed{\bf{\orange{RHS:- 3}}}}$

$\Large\underline\bold\red{Some\: important\: identities}$

• (a + b)2 = a2 + 2ab + b2
• (a – b)2 = a2 – 2ab + b2
• a2 – b2= (a + b)(a – b)
• (x + a)(x + b) = x2 + (a + b) x + ab
• (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
• (a + b)3 = a3 + b3 + 3ab (a + b)
• (a – b)3 = a3 – b3 – 3ab (a – b)
• a3 + b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2 – ab – bc – ca)

Answer 2

QUESTION:-

$\sf\therefore \dfrac{(b+c)^2}{bc} + \dfrac{(c+a)^2}{ca} + \dfrac{(a+b)^2}{ab}$

$\sf\large\therefore note:-question\:is\:incorrect$

CORRECT QUESTION,

$\sf{\text{If \:a+b+c =0\: then\: show ,}}$

$\bf\therefore \frac{(b+c)^2}{bc} + \frac{(c+a)^2}{ca} + \frac{(a+b)^2}{ab}$

ANSWER✔

$\Large\underline\bold{GIVEN,}$

$\sf\implies a + b + c=0.$

$\sf\implies a + b = -c$

$\sf\implies a + c = -a$

$\sf\implies c + a = -1$

$\Large\underline\bold{TO\:PROVE,}$

$\sf\large\therefore \frac{(b+c)^2}{bc} + \frac{(c+a)^2}{ca} + \frac{(a+b)^2}{ab}$

$\Large\underline\bold{SOLUTION,}$

$\sf\therefore Taking \:LHS,$

$\sf\large\implies \frac{(b+c)^2}{bc} + \frac{(c+a)^2}{ca} + \frac{(a+b)^2}{ab}$

$\sf\large\therefore substituting \:the \:value\: of \:eq. \:1,2,3$

$\sf\large\implies \frac{(-a)^2}{bc} + \frac{(-b)^2}{ca} + \frac{(-c)^2}{ab}$

$\sf\large\implies \frac{a^2}{bc} + \frac{b^2}{ca} + \frac{c^2}{ab}$

$\sf\therefore taking\:LCM,$

$\sf\large\implies \frac{a^3+b^3+c^3}{abc}$

$\sf\therefore we\:know$

$\sf\large\implies (a+b+c)=0 , a^3+b^3+c^3=3abc$

$\sf\large\implies \frac{3abc}{abc}$

$\sf\large\implies RHS=3$

$\large{\boxed{\sf{RHS=3}}}$

HENCE,R.H.S =3