Question

if a+b+c=o then a^2/bc+b^2/ca+c^2/ab

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Answer 1

Given :

• If $\\ \rm{a+b+c=0}$

•then $\\ \rm{\dfrac{a^{2}}{bc}+\dfrac{b^{2}}{ca}+\dfrac{c^{2}}{ab}}$

To Find :

$\\ \boxed{\sf{The \: value \: of \: \dfrac{a^{2}}{bc}+\dfrac{b^{2}}{ca}+\dfrac{c^{2}}{ab}=?}}$

Formulas :

$\pink \bigstar \bf \: {a}^{3} + {b}^{3} + {c}^{3} = (a + b + c)( {a}^{2} + {b}^{2} + {c}^{2} - ab - bc - ca) + 3 abc$

Solution:

Given equation,

➡ a+b+c=0

we have,

$\tt \bigstar {a}^{3} + {b}^{3} + {c}^{3} = (a + b + c)( {a}^{2} + {b}^{2} + {c}^{2} - ab - bc - ca) + 3abc$

If a+b+c=0

then

$\\ \implies \sf \: {a}^{3} + {b}^{3} + {c}^{3} = (0)( {a}^{2} + {b}^{2} + {c}^{2} - ab - bc - ca) + 3abc \\ \\ \implies \sf \: {a}^{3} + {b}^{3} + {c}^{3} = 3abc \: \: \: \qquad.....equation(1)$

$\\ \implies \sf \: \frac{ {a}^{2} }{bc} + \frac{ {b}^{2} }{ca} + \frac{ {c}^{2} }{ab} \\ \\ \implies \bold{ \red{ \underline{multipying \: both \: sides \: a \: b \: and \: c}}} \\ \\ \\ \implies \sf \: \frac{ {a}^{2} }{bc} \times \frac{a}{a} + \frac{ {b}^{2} }{ac} \times \frac{b}{b} + \frac{ {c}^{2} }{ab} \times \frac{c}{c} \\ \\ \\ \implies \sf \: \frac{ {a}^{3} }{abc} + \frac{ {b}^{3} }{abc} + \frac{ {c}^{3} }{abc} \\ \\ \\ \implies \sf \: \frac{ {a}^{3} + {b}^{3} + {c}^{3} }{abc} \\ \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \implies \sf \quad \ \frac{3abc}{abc} \: \: \: \qquad \: ...from \: equation(1) \\ \\ \\ \implies \sf \: 3 \\ \\ \\ \boxed{ \pink \bigstar{ \bf{ \frac{ {a}^{2} }{bc} + \frac{ {b}^{2} }{ac} + \frac{ {c}^{2} }{ab} = 3}}}$

➡ The Correct answer will be 3.