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
