Question

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

Answer 1

a + b + c = 0

From this relation :

b + c = - a

c + a = -b

a + b = - c

a + b = -c

cubing on both sides

(a + b)³ = (-c)³

a³ + b³ + 3ab(a + b) = -c³

a³ + b³ + 3ab(-c) = -c³

a³ + b³ + c³ = 3abc

To prove:

a(b+c)² + b(c+a)² + c(a+b)² = 3abc

Taking L.H.S

a(-a)² + b(-b)² + c(-c)² = 3abc

a³ + b³ + c³ = 3abc

3abc = 3abc [ since a³ + b³ + c³ = 3abc]

Hence proved.