if a+b+c =0 then prove that a2/ab+b2/bc+c2/ca= 3
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Solution :
It is given that ,
a + b + c = 0 ----( 1 )
a³ + b³ + c³ = 3abc ----( 2 )
LHS = a²/bc + b²/ac + c²/ab
= ( a³ + b³ + c³ )/abc
= 3abc/abc [ from ( 1 ) ]
= 3
= RHS
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