21Let a(b-c)=x
b(c-a)=y
c(a-b)=z
x+y+z= 0
so x3 +y3+z3=3xyz
a3(b-c)3 +b3(c-a)3 +c3(a-b)3=3abc(a-b)(b-c)(c-a)
2 Answers
21
62Another method would be to observe that
if a=0, then the given function is zero
if b=0 then it is zero
c=0 then it is zero..
also if a=b then function is zero and so on
So a, b, c, a-b, b-c, c-a will all be the factors of the given experession
so the function will be
k. abc(a-b)(b-c)(c-a)
substitute a=1, b=-1 and c=2 to find the value of the constant. :)