Question

If a+b+c=0. Prove a^2÷bc+b^2÷ca+c^2+ab=3​

Answer 1

Given :-

$\sf{(a + b + c) = 0 \: }$

To Prove :-

$\sf{\dfrac{a^{2}}{bc} + \dfrac{b^{2}}{ca} + \dfrac{c^{2}}{ab} = 3}$

Using Identity :-

If , $\sf{(a + b + c) = 0 \: }$

then , $\sf{(a^{3} + b^{3} + c^{3}) = 3abc}$

Solution :-

From L . H . S -

$\sf{\dfrac{a^{2}}{bc} + \dfrac{b^{2}}{ac} + \dfrac{c^{2}}{ab}}$

$\sf{= \dfrac{a^{2}\times a}{bc\times a} + \dfrac{b^{2}\times b}{ac\times b} + \dfrac{c^{2}\times c}{ab\times c}}$

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

$\sf{= \dfrac{(a^{3} + b^{3} + c^{3})}{abc}}$

By Using the identity -

$\sf{= \: \dfrac{3abc}{abc}}$

$\sf{= \: 3}$

$\sf{= R.H.S.}$

HENCE PROOF

★ Extra Using Identity :-

$\sf{a^{3} + b^{3} + c^{3} - 3abc = (a + b + c)[(a^{2} + b^{2} + c^{2}) - (ab + bc + ca)]}$