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