If $$a^{3}-b^{3}-c^{3}=0$$ then the value of $$a^{9}-b^{9}-c^{9}-3a^{3} b^{3} c^{3}$$ is

shortcut : 

put c = 0 in  $$a^{3}-b^{3}-c^{3}=0$$ $$\Rightarrow$$ $$a^{3}=b^{3}$$

$$a^{9}-b^{9}-(0)^{9}-3a^{3} b^{3} (0)^{3}$$ = $$a^{9}-b^{9}$$ = $$(a^{3})^{3}-(b^{3})^{3}$$ =  $$(a)^{3}-(a)^{3}$$ = 0  ( $$\because$$ $$a^{3}=b^{3}$$ )

 so the answer is option C.

normal method : 

$$a^{3}-b^{3}-c^{3}=0$$

$$a^{3}=b^{3}+c^{3}$$

cubing on both sides, 

$$(a^{3})^{3}=(b^{3}+c^{3})^{3}$$

$$a^{9}=b^{9}+c^{9}+3b^{3} c^{3}(b^{3}+c^{3})$$

$$a^{9}=b^{9}+c^{9}+3b^{3} c^{3}(a^{3})$$

$$a^{9}-b^{9}-c^{9}-3a^{3}b^{3} c^{3}=0$$

so the answer is option C.