Question

If a,b,c are all non zero and a+b+c=0,prove that a²/bc +b²/ca +c²/ab=3​

Answer 1

Given that,

a+b+c=0

To prove that:

a²/bc+b²/ac+c²/ab

★Using the following identity,

When x+y+z=0,then x³+y³+z³=3xyz

Now,

a²/bc+b²/ca+c²/ab

Taking LCM,

→a³+b³+c³/abc

★We know that,

a³+b³+c³=3abc

On substitution of the value,we obtain:

→3abc/abc

→3

Thus,a²/bc+b²/ac+c²/ab=3

Hence,proved

Answer 2

Solution:

Given that ;

a, b and c are all non-zero and a + b + c = 0.

To prove : $\sf \dfrac{a^2}{bc} + \dfrac{b^2}{ca} + \dfrac{c^2}{ab}$

Proof :

We know that ;

When, x + y + z = 0, then x³ + y³ + z³ = 3abc. We will use this identity to prove the above statement. Here we go ;

$\sf \frac{a^2}{bc} + \frac{b^2}{ca} + \frac{c^2}{ab} \\ \\ \bf Taking \: LCM : \\ \\ \sf \implies \frac{a {}^{3} + b {}^{3} + c {}^{3} }{abc} \\ \\ \sf \implies \frac{3abc}{abc}$

Since, a + b + c= 0, so a³ + b³ + c³ = 3abc.

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

Hence, it is proved.