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 :Â
put c = 0 in  $$a^{3}-b^{3}-c^{3}=0$$ $$\Rightarrow$$ $$a^{3}=b^{3}$$
$$a^{9}-b^{9}-(0)^{9}-3a^{3} b^{3} (0)^{3}$$ =Â $$a^{9}-b^{9}$$ =Â $$(a^{3})^{3}-(b^{3})^{3}$$ =Â Â $$(a)^{3}-(a)^{3}$$ = 0Â ( $$\because$$ $$a^{3}=b^{3}$$ )
 so the answer is option C.
normal method :Â
$$a^{3}-b^{3}-c^{3}=0$$
$$a^{3}=b^{3}+c^{3}$$
cubing on both sides,Â
$$(a^{3})^{3}=(b^{3}+c^{3})^{3}$$
$$a^{9}=b^{9}+c^{9}+3b^{3} c^{3}(b^{3}+c^{3})$$
$$a^{9}=b^{9}+c^{9}+3b^{3} c^{3}(a^{3})$$
$$a^{9}-b^{9}-c^{9}-3a^{3}b^{3} c^{3}=0$$
so the answer is option C.
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