Question

If a,b,c are all non-zero numbers and a+b+c=0, prove that


`a ^(2) / bc + b²/ac + c²/ab = 3`
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Answer 1

Given - a+b+c = 0

To prove that-

a²/bc + b²/ac + c²/ab = 3

Now we know that

when x+y+z = 0,

then x³+y³+z³ = 3xyz

that means

(x³+y³+z³)/xyz = 3 ---- eq 1)

Lets solve for LHS

LHS = a²/bc + b²/ac + c²/ab

we can write it as LHS = a³/abc + b³/abc + c³/abc

by multiplying missing denominators,

now take common abc from denominator and you'll get,

LHS = (a³+b³+c³)/abc --- eq (2)

Comparing one and two we can say that

(a³+b³+c³)/abc = 3

Hence proved,

a²/bc + b²/ac + c²/ab = 3