Question

If a+b+c=0 prove that
$\frac{ {a}^{2} }{bc} + \frac{ {b}^{2} }{ab} + \frac{ {c}^{2} }{ac} = 3$
​

Answer 1

Answer:

$as \: a + b + c = 0$

${a}^{3} + {b}^{3} + {c}^{3} = 0$

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

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

$= > 3$

Answer 2

Answer:

$\huge\bold\red{ANSWER}$

Step-by-step explanation:

$\large{\underline{\rm{\red{SOLUTION :}}}}$

$\implies \sf \blue{if\:a+b+c=0}\\ \sf \blue {S.O.B.S}\\ \sf \blue {a^2+b^2+c^2+2ab+2bc+2ca=0}\\ \implies \sf \pink {\frac{a^2}{bc}+\frac{b^2}{ab}+\frac{c^2}{ac}=3}\\ \implies \sf \pink {\frac{a^3+b^3+c^3}{abc}=3}\\ \implies \sf \pink {a^3+b^3+c^3=3abc}\\ \sf {this \:occurs\:by\:cubing\:a+b+c=0}$