Correct Answer: $\frac{3xyz}{abc}$

Solution : Given: $x=a\left (b-c \right),\; y=b\left (c-a \right) ,\; z=c\left (a-b \right)$
$\left (\frac{x}{a} \right )^{3}+\left (\frac{y}{b} \right )^{3}+\left (\frac{z}{c} \right)^{3}$
= $ (b-c)^3+(c-a)^3+(a-b)^3$
Since $\left (\frac{x}{a} \right )+\left (\frac{y}{b} \right )+\left (\frac{z}{c} \right)$
= $ (b-c)+(c-a)+(a-b)=0$
Use identity: $x^3+y^3+z^3=3xyz$ if $x+y+z=0$
So, $\left (\frac{x}{a} \right )^{3}+\left (\frac{y}{b} \right )^{3}+\left (\frac{z}{c} \right)^{3}$
= $3 \left (\frac{x}{a} \right )(\frac{y}{b})(\frac{z}{c})=\frac{3xyz}{abc}$
Hence, the correct answer is $\frac{3xyz}{abc}$.