Question

if a+b+c=0, find the value of
(b+c)²/bc+(c+a)²/ca+(a+b)²/ab​

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Answer:

3

Step-by-step explanation:

Given : a + b + c = 0

To Find : $\dfrac{ {(b + c)}^{2} }{bc} + \dfrac{ {(c + a)}^{2} }{ca} + \dfrac{ {(a + b)}^{2} }{ab}$

From the given equation, we can find out :

a + b + c = 0

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

Putting these equations in the above equation, we get

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

→ $\dfrac{ {a}^{2} }{bc} + \dfrac{ {b}^{2} }{ca} + \dfrac{ {c}^{2} }{ab}$

→ $\dfrac{a {(a)}^{2} + b {(b)}^{2} + c {(c)}^{2} }{abc}$

→ $\dfrac{ {a}^{3} + {b}^{3} + {c}^{3} }{abc}$

• {Identity : (a + b + c)³ = a³ + b³ + c³ - 3abc }

→ Given that a + b + c = 0, then

(0)³ = a³ + b³ + c³ - 3abc

a³ + b³ + c³ = 3abc

Putting this in above equation, we get

→ $\dfrac{3abc}{abc}$

→ 3