Question

hey frds plz help

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Answer 1

Question:

$Find \: the \: value \: of : \frac{ {(a + b)}^{2} }{ab} + \frac{ {(b + c)}^{2} }{bc} + \frac{ {(c + a)}^{2} }{ca}$

Answer:

$The \: value \: is \: \boxed{3}$

Step-by-step explanation:

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

Rule :-

In this case we know that if a + b + c = 0 ,

then a³ + b³ + c³ = 0 .

$a + b + c = 0$

$\boxed{a + b = - c}$

$\boxed{b + c = - a}$

$\boxed{c + a = - b}$

$= \frac{{( - c)}^{2} }{ab}+\frac{{( - a)}^{2} }{bc}+ \frac{{( - b)}^{2}}{ca}$

$= \frac{{( c)}^{2} }{ab}+\frac{{( a)}^{2} }{bc}+ \frac{{( b)}^{2}}{ca}$

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

L.C.M. is abc

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

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

$\boxed{= 3}$

Formula & Rules :-

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

Here, a + b + c = 0

${a}^{3} + {b}^{3} + {c}^{3} = (0)( {a}^{2} + {b}^{2} + {c}^{2} - ab - bc - ca) + (3abc)$

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