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

Solution(By Examveda Team)

Given x = a (b - c), y = b (c - a), z = c (a - b)
x = a (b - c)
⇒ $$\frac{x}{a}$$ = b - c ..... (i)
Similarly, y = b (c - a)
⇒ $$\frac{y}{b}$$ = c - a ..... (ii)
And similarly z = c (a - b)
⇒ $$\frac{z}{c}$$ = c - a ..... (iii)
Adding (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
$$\eqalign{ & \therefore {\left( {\frac{x}{a}} \right)^3} + {\left( {\frac{y}{b}} \right)^3} + {\left( {\frac{z}{c}} \right)^3} \cr & = 3 \times \frac{x}{a} \times \frac{y}{b} \times \frac{z}{c} \cr & = \frac{{3xyz}}{{abc}} \cr} $$
[If a + b + c = 0, a³ + b³ + c³ = 3abc]