23. If a + b + c = 0, prove that
a2/bc+b2/ca+c2/ab= 3
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Given,
a+b+c = 0
=>(a+b+c)³ = 0³
=>(a+b)³+c³+3(a+b)(c)(a+b+c) = 0
=>a³+b³+3ab(a+b)+c³+3(a+b)(c)(0) = 0
=>a³+b³+3ab(a+b+c-c)+c³+0 = 0
=>a³+b³+3ab(0-c)+c³ = 0
=>a³+b³+c³-3abc = 0
=>a³+b³+c³ = 3abc
=>(a³/abc)+(b³/abc)+(c³/abc) = 3
=>(a²/bc)+(b²/ac)+(c²/ab) = 3 (Proved)
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