Question

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

Answer 1

hey mate here's ur ans ↓↓

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Let, x = a-b , y = b-c , z = c - a

∴ x + y+ z = (a - b) + ( b - c) + (c - a) = 0

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

Now,we know that ,

( x + y + z)³ = x³ + y³+ z³ + 3( x+y)(y+z)(z+x)
⇒0 =( a - b)³ + (b - c)³ + (c - a )³ + 3(a - c)(b-a)(c-b)
⇒(a - b)³ + (b - c)³ + (c - a)³ = -3 (a -c)( b - a)(c -b)
⇒(a - b)³ + (b - c)³ + (c - a)³ = 3(a -b )( b - c)(c - a)

∴Proved.

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hope it helps u☺☺✌

Answer 2

Answer:

Let,

a-b=x

b-c=y

c-a=z

x+y+z=0

=X2+y3+z3=3xyz

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

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