The answer to the above equation is 3
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