Math, asked by atanukar10oy0a4k, 11 months ago

if a+b+c=0 then show that a^2(b+c)+b^2(c+a)+c^2(a+b)+3abc=0​

Answers

Answered by manasvi1604
21

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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