Question

Find the value
(a-b)^3+(b-c)^3+(c-a)^3/3(a+b) (b+c) (c+a)

Answer 1

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

Here observe that

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

and we know that if x + y + z = 0, then

x³ + y³ + z³ =3xyz

Putting back values of x, y and z

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

We need to find$\frac{(a-b)^3 + (b-c)^3 +(c-a)^3}{3(a+b)(b+c)(c+a)}$

$\frac{3(a-b)(b-c)(c-a)}{3(a+b)(b+c)(c+a)}$

\frac{(a-b)(b-c)(c-a)}{(a+b)(b+c)(c+a)} \\