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## Answers

**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)**

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A³ (b - c)³ + b³ ( c - a)³ + c³ (a - b)³

= 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) ]

= - 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 ]

*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)*