# Factorise a^3(b-c)^3+b^3(c-a)^3+c^3(a-b)^3

2
by aahtews

2015-04-25T18:05:11+05:30
a^3(b-c)^3+b^3(c-a)^3+c^3(a-b)^3 = [a(b-c)]^3 + [b(c-a)]^3 + [c(a-b)]^3
We know, a^3 + b^3 + c^3 =3abc if a+b+c =0
Here, a =
[a(b-c)] , b = [b(c-a)] & c =[c(a-b)]
⇒a+b+c =  [a(b-c)] + [b(c-a)] + [c(a-b)]
= ab - ac + bc - ab + ac - bc = 0
Therefore, a^3 + b^3 + c^3 = 3abc
⇒[a(b-c)]^3 + [b(c-a)]^3 + [c(a-b)]^3 = 3 [a(b-c)] [b(c-a)] [c(a-b)]
= 3abc* (b-c)* (c-a)* (a-b)
Thank u but the line 2 contains an error it should have been a^3+b^3+c^3=3abc if a+b+c=0
2015-04-25T23:40:23+05:30

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A³ (b - c)³ + b³ ( c - a)³ + c³ (a - b)³
expand the cubes and cancel the equal and opposite terms.

= a³ b³ - a³ c³ - 3 a³ b c (b - c)    +    b³ c³ - b³ a³ - 3 b³ a c (c - a)
+ c³ a³ - c³ b³ - 3 c³ a b (a - b)

= - 3 a b c [ a² (b - c) + b² (c - a) + c² (a - b) ]
write  b - c  as  b - a  +  a - c

=  - 3 a b c [ a² (b - a)  + a² (a - c) + b² (c - a) + c² (a - b) ]
= -  3 a b c  [  (a - b) (c² - a²) + (c - a) (b² - a²) ]
= -  3 a b c  [ (a - b) (c + a) ( c - a) + (c -a )  ( b+ a) ( b - a) ]
=  - 3 a b c    (a - b) (c - a) [ c + a - b - a ]
=  3 a b c (a - b) (c - a)  (b - c)