Question

If a+b+c=0 then show that (b+c)^2/bc + ( c+a)^2/ca +(a+b)^2/ab=3

Answer 1

||✪✪ QUESTION ✪✪||

If a+b+c=0 then show that (b+c)^2/bc + ( c+a)^2/ca +(a+b)^2/ab=3

|| ✰✰ ANSWER ✰✰ ||

Given,

a +b + c = 0

→ a + b = -c ----- Equation (1)

→ b + c = -a ---- Equation (2)

→ c + a = -b ---- Equation (3)

Now, To Prove :- (b+c)^2/bc + ( c+a)^2/ca +(a+b)^2/ab = 3

Taking LHS,

→ (b+c)^2/bc + ( c+a)^2/ca +(a+b)^2/ab

Putting value of Eq (1), Eq.(2) and Eq.(3) ,,

→ (-a)²/bc + (-b)²/ca + (-c)²/ab

→ a²/bc + b²/ca + c²/ab

Taking LCM now,

→ (a³ + b³ + c³)/abc

Now, we know That, if (a+b+c) = 0, a³ + b³ + c³ = 3abc ,

So ,

→ 3abc/abc

→ 3 = RHS.

✪✪ Hence Proved ✪✪

Answer 2

$\bf Given:\ \sf a\ +\ b\ +\ c\ =\ 0.\\\\\\\bf To\ prove\ that:\ \sf \dfrac{(b\ +\ c)^2}{bc}\ +\ \dfrac{(c\ +\ a)^2}{ca}\ +\ \dfrac{(a\ +\ b)^2}{ab}\ =\ 3.\\\\\\\bf Answer:\\\\\\\sf From\ the\ given\ data,\ we\ can\ say\ that:\\\\\\\ \tt a\ +\ b\ =\ -\ c\\\\b\ +\ c\ =\ -\ a\\\\c\ +\ a\ =\ -\ b\\\\\\\bf Left\ hand\ side,\\\\\\\sf \dfrac{(b\ +\ c)^2}{bc}\ +\ \dfrac{(c\ +\ a)^2}{ca}\ +\ \dfrac{(a\ +\ b)^2}{ab}\\\\\\\bf Using\ the\ values\ we\ derived,$

$\sf \dfrac{(-\ a)^2}{bc}\ +\ \dfrac{(-\ b)^2}{ca}\ +\ \dfrac{(-\ c)^2}{ab}\\\\\\=\ \dfrac{a^2}{bc}\ +\ \dfrac{b^2}{ca}\ +\ \dfrac{c^2}{ab}\\\\\\=\ \dfrac{a^3\ +\ b^3\ +\ c^3}{abc}\\\\\\\bf We\ know\ that\ if\ a\ +\ b\ +\ c\ =\ 0,\ then\ a^3\ +\ b^3\ +\ c^3\ =\ 3abc.\\\\\\According\ to\ the\ question,\ a\ +\ b\ +\ c\ =\ 0.\\\\\\\sf =\ \dfrac{3abc}{abc}\\\\\\= 3\\\\=\ \bf Right\ hand\ side.$