If $$x=a(b-c),y=b(c-a)$$​ and $$z=c(a-b)$$​ then what will be the value of $$(\frac{x}{a})^3+(\frac{y}{b})^3+(\frac{z}{c})^3$$​ ?

Correct option is D

Given:

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

$$\left (\frac{x}{a}\right )^3+\left (\frac{y}{b}\right)^3+\left (\frac{z}{c}\right)^3$$= ?

Concept Used:

a³ + b³ + c³ - 3abc = (a + b + c) (a² + b² + c² - ab - bc - ca)

If a + b + c = 0 then, a³ + b³ + c³ = 3abc

Solution:​

​x = a(b - c)

$$\frac xa$$ = b - c     ....(1)

y = b(c - a)

$$\frac yb$$ = c - a     .....(2)

z = c(a - b)

$$\frac zc$$ = a - b      ....(3)

Adding equations 1, 2 and 3, we have -

$$\frac xa +\frac yb +\frac zc=$$ b - c + c - a + a - b = 0

Now, $$\left (\frac{x}{a}\right )^3+\left (\frac{y}{b}\right)^3+\left (\frac{z}{c}\right)^3$$

=$$\frac{3xyz}{abc}$$​