Question

Please... it's urgent...please do the answer..If a+b+c=0,find the value of a²/bc + b²/ac +c²/ab.

Answer 1

Answer:

3

Step-by-step explanation:

Given that ;

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

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

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, the answer is 3.

Answer 2

Answer:

3

Step-by-step explanation:

Given:

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

$\huge{\frac{a^{2}}{bc}}$ + $\huge{\frac{b^{2}}{ca}}$ + $\huge{\frac{c^{2}}{ab}}$

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;

$\huge{\frac{a^{2}}{bc}}$ + $\huge{\frac{b^{2}}{ca}}$ + $\huge{\frac{c^{2}}{ab}}$

Taking LCM:

= $\huge{\frac{a^{3} + b^{3} + c^{3}}{abc}}$

= $\huge{\frac{3abc}{abc}}$

Since, a + b + c = 0

So, a³ + b³ + c³ = 3abc

= $\huge{\frac{3abc}{abc}}$

= 3

Hope it is helpful....