Question

show that (a-b)cube+(b-c)cube+(c-a)cube= 3(a-b)(a+b)(c-a)​

Answer 1

let take x=(a-b), y=(b-c), z=(c-a)

therefore, x+y+z=(a-b)+(b-c)+(c-a)

so x+y+z=0, as a, b, c get cancelled out.

now (x+y+z)³=x³+y³+z³+3xyz (it's a property)

since, x+y+z=0

therefore, x³+y³+z³+3xyz=0

now put value of x, y and z

(a-b)³+(b-c)³+(c-a)³+3(a-b)(b-c)(c-a)=0

(a-b)³+(b-c)³+(c-a)³= -3(a-b)(b-c)(c-a)

(a-b)³+(b-c)³+(c-a)³=3(a-b)(b+c)(c-a)

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