if a+b+c=0,then evaluate a^2/bc+b^2/ca+c^2/ab. solve stepwise
adamsyakir:
not complete ?
it is a complete question
oh sorry at the end of the equation it will not be ab but abc
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a²/bc+b²/ca+c²/ab
=(a³+b³+c³)/abc
={(a³+b³)+c³}/abc
=[{(a+b)³-3ab(a+b)}+c³]/abc
=[{(-c)³-3ab(-c)}+c³]/abc [∵, a+b+c=0; ∴, a+b=-c]
=(-c³+3abc+c³)/abc
=3abc/abc
=3
=(a³+b³+c³)/abc
={(a³+b³)+c³}/abc
=[{(a+b)³-3ab(a+b)}+c³]/abc
=[{(-c)³-3ab(-c)}+c³]/abc [∵, a+b+c=0; ∴, a+b=-c]
=(-c³+3abc+c³)/abc
=3abc/abc
=3
awesome answer thanks a lot
welcome :)
cyclic factor formula, a3 + b3 +c3 - 3abc = ( a + b + c ) ( a2 + b2 + c2 - ab - bc = ca ). if a + b + c = 0 then a3 + b3 + c3 = 3abc, hence a3 + b3 + c3 / abc = 3abc / abc = 3
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