If x=a(b-c), y=b(c-a), z=c(a-b), then the value of $$(\frac{x}{a})^3\ +\ (\frac{y}{b})^3\ +\ (\frac{z}{c})^3\ $$is:

Given : $$x=a(b-c)$$ , $$y=b(c-a)$$ , $$z=c(a-b)$$

=> $$\frac{x}{a}=(b-c)$$ -------------(i)

and $$\frac{y}{b}=(c-a)$$ -------------(ii)

and $$\frac{z}{c}=(a-b)$$ -------------(iii)

Adding equations (i), (ii) and (iii), we get :

=> $$\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=(b-c)+(c-a)+(a-b)$$

=> $$\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=0$$

Now, we know that if $$(p+q+r)=0$$, then $$p^3+q^3+r^3=3pqr$$

$$\therefore$$ $$(\frac{x}{a})^3\ +\ (\frac{y}{b})^3\ +\ (\frac{z}{c})^3\ $$

= $$3\times(\frac{x}{a})\times(\frac{y}{b})\times(\frac{z}{c})$$

= $$\frac{3xyz}{abc}$$

=> Ans - (C)