if a+b+c=0 then show that a^2(b+c)+b^2(c+a)+c^2(a+b)+3abc=0
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Answer:
a^2(b+c)+b^2(c+a)+c^2(a+b)+3abc=0
L.H.S
a^2b + a^2c + b^2a + c^2a + c^2b + 3abc
a^2b + a^2c + b^2a + c^2a + 3abc + c^2b + b^2c
a(ab + ac + b^2 + c^2+ 3bc) + bc( c+b)
a(ab + ac + b^2 + c^2 + 2bc +bc) + bc( c+b)
a{ ab + ac + bc + (b+c)^2} +bc(b+c)
a{ ab + ac + bc + (-a)^2} +bc(b+c)
a{ ab + ac + bc + a^2} +bc(b+c)
a{ a(a+b+c) + bc} +bc(b+c)
a{ a(0) + bc} +bc(b+c)
a(bc) + bc(b+c)
bc ( a+b+c )
bc(0)
0
HENCE PROVED
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