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)

$$\eqalign{ & x = a\left( {b - c} \right) \cr & y = b\left( {c - a} \right) \cr & z = c\left( {a - b} \right) \cr & {\text{Let, }} \cr & \frac{x}{a} = b - c{\text{ }}\,\,\,\,\,\,\,\,\frac{x}{a} = {\text{A}} \cr & \frac{y}{b} = c - a{\text{ }}\,\,\,\,\,\,\,\,\frac{y}{b} = {\text{B}} \cr & \frac{z}{c} = a - b{\text{ }}\,\,\,\,\,\,\,\,\frac{z}{c} = {\text{C}} \cr & \therefore {\text{A}} + {\text{B}} + {\text{C}} \cr & = b - c + c - a + a - b \cr & = 0 \cr & \therefore {{\text{A}}^3} + {{\text{B}}^3} + {{\text{C}}^3} = {\text{3ABC}} \cr & \therefore {\left( {\frac{x}{a}} \right)^3}{\text{ + }}{\left( {\frac{y}{b}} \right)^3} + {\left( {\frac{z}{c}} \right)^3} \cr & = 3 \times \frac{x}{a} \times \frac{y}{b} \times \frac{z}{c} \cr & = \frac{{3xyz}}{{abc}} \cr} $$