A-b) 3 + (b-c) 3 + (c-a) 3: (a-b) (b-c) (c-a)

The answer to the above equation is 3

Step-by-step explanation:

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

Let us consider (a-b) = x, (b-c) = y and (c-a) = z.

Hence, It is obvious that:

x+y+z = 0 ∵all the terms gets cancelled out

⇒We must remember the algebraic formula

x³+y³+z³-3xyz = (x+y+z) (x²+y²+z²-xy-xz-yz)

Since x+y+z=0 ⇒Whole " (x+y+z) (x²+y²+z²-xy-xz-yz) " term becomes 0

x³+y³+z³-3xyz = 0

Alternatively, x³+y³+z³ = 3xyz

Now putting the value of x, y, z in the original equation

(a-b) ³ + (b-c) ³ + (c-a) ³ can be written as 3 (a-b) (b-c) (c-a) since (a-b) = x, (b-c) = y and (c-a) = z.

3 (a-b) (b-c) (c-a) : (a-b) (b-c) (c-a)

= 3 ∵Common factor (a-b) (b-c) (c-a) gets cancelled out

Answer to the above question is 3