Question

Ifa+b+c=0, find the value of
(b+c)^2÷bc+(c+a)^2÷ca+(a+b)^2÷ab​

Answer 1

Answer:

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

Step-by-step explanation:

Given,

a + b+ c = 0

We know that,

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

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

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

To find the value of,

= $\frac{(b+c)^{2} }{bc} + \frac{(a+c)^{2} }{ac} + \frac{(a+b )^{2} }{ab}$

Solution:

Expanding using $(a+b)^{2} = (a^{2} +b^{2} + 2ab )$ formula,

=$\frac{(a^{2} + b^{2} + 2ab) }{ab} + \frac{(b^{2} + c^{2} + 2bc) }{bc} +\frac{(a^{2} + c^{2} + 2ac) }{ac}$

= $(\frac{a^{2} }{ab} + \frac{b^{2} }{ab} +\frac{2ab}{ab} }) + (\frac{b^{2} }{bc} + \frac{c^{2} }{bc} +\frac{2bc}{bc} }) + (\frac{a^{2} }{ac} + \frac{c^{2} }{ac} +\frac{2ac}{ac} })$

Eliminating common terms,

= $(\frac{a}{b}+\frac{b}{a} + 2) + (\frac{b}{c}+\frac{c}{b} + 2) + (\frac{a}{c}+\frac{c}{a} + 2)$

= $(\frac{b}{a} +\frac{c}{a}+ \frac{a}{b} +\frac{a}{c} +\frac{b}{c} + 2+ 2 + 2)$

=$\frac{(b+c)}{a} + \frac{(a+c)}{b} +\frac{(a+b)}{c} + 6$

Putting (1),(2),(3) in the above equation,

=$\frac{(-a)}{a} + \frac{(-b)}{b} +\frac{(-c)}{c} + 6$

= (-1) +(-1) +(-1) + 6

= 6 - 3

= 3

Hence,

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

Hope it helps...