1. 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
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By: anil on 05 May 2019 01.55 am
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 ( $$ecause$$ $$a^{3}=b^{3}$$ ) 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.
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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.