Question

a) If a+b+c=0 then show that a²(b+c)+b²(c+a)+c²(a+b)+3abc=0.



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Answer 1

Step-by-step explanation:

a+b+c=0

a+b= -c

b+c = -a

c+a= -b

a²(b+c)+b²(c+a)+c²(a+b)+3abc

a²(-a) + b²(-b) + c²(-c) +3abc

-a³-b³-c³+3abc

Take - sign common

-(a³+b³+c³-3abc)

We know that a³+b³+c³-3abc= (a+b+c)(a²+b²+c²-ab-bc-ca)

a+b+c=0(Given)

-(a³+b³+c³-3abc)=(0)(a²+b²+c²-ab-bc-ca)

a³+b³+c³-3abc=0

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