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Category: Algebra

Question-163957

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Question Number 163957 by HongKing last updated on 12/Jan/22 Answered by mahdipoor last updated on 12/Jan/22 $${get}\:{n}=\mathrm{3}{b}+{a}\:,\:{b}\in{N}\:\:\:\:\mathrm{0}\leqslant{a}<\mathrm{3} \\ $$$$\left[{n}\right]=\mathrm{3}{b}+\left[{a}\right]\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left[\frac{{n}+\mathrm{1}}{\mathrm{3}}\right]={b}+\left[\frac{{a}+\mathrm{1}}{\mathrm{3}}\right] \\ $$$$\left[\frac{{n}}{\mathrm{3}}\right]={b}+\left[\frac{{a}}{\mathrm{3}}\right]={b}\:\:\:\:\:\:\:\:\:\:\left[\frac{{n}+\mathrm{2}}{\mathrm{3}}\right]={b}+\left[\frac{{a}+\mathrm{2}}{\mathrm{3}}\right] \\ $$$$\Rightarrow\left(−\mathrm{1}\right)^{\mathrm{3}{b}+\left[{a}\right]} +\left(−\mathrm{1}\right)^{{b}+\left[\frac{{a}+\mathrm{1}}{\mathrm{3}}\right]} =\left(−\mathrm{1}\right)^{{b}}…

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Question-98416

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Question Number 98416 by 175 last updated on 13/Jun/20 Answered by Farruxjano last updated on 13/Jun/20 $$\boldsymbol{{i}}^{\mathrm{5}\boldsymbol{{n}}+\mathrm{1}} =\mathrm{1}\:\Rightarrow\:\mathrm{5}\boldsymbol{{n}}+\mathrm{1}=\mathrm{2}\boldsymbol{{k}}\:\left(\boldsymbol{{k}}\in\boldsymbol{{Z}}\right),\:\Rightarrow\:\boldsymbol{{n}}=\mathrm{2}\boldsymbol{{t}}+\mathrm{1}\:\left(\boldsymbol{{t}}\in\boldsymbol{{Z}}\right) \\ $$ Answered by Ramajunan last updated…

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x-3-x-6-8y-y-3-y-6-8z-z-3-z-6-8x-x-y-z-

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Question Number 163906 by HongKing last updated on 11/Jan/22 $$\begin{cases}{\mathrm{x}^{\mathrm{3}} \:+\:\mathrm{x}\:+\:\mathrm{6}\:=\:\mathrm{8y}}\\{\mathrm{y}^{\mathrm{3}} \:+\:\mathrm{y}\:+\:\mathrm{6}\:=\:\mathrm{8z}}\\{\mathrm{z}^{\mathrm{3}} \:+\:\mathrm{z}\:+\:\mathrm{6}\:=\:\mathrm{8x}}\end{cases}\:\:\:\Rightarrow\:\:\:\mathrm{x};\mathrm{y};\mathrm{z}\:=\:? \\ $$ Answered by ajfour last updated on 11/Jan/22 $${let}\:\:{from}\:{symmetry} \\ $$$${x}={y}={z}\:\:\Rightarrow\:\:{x}^{\mathrm{3}}…

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