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Category: Algebra

2-a-2-b-2-c-x-Find-natural-numbers-a-b-c-such-that-the-number-x-is-a-cube-of-any-number-

Question Number 145279 by mathdanisur last updated on 03/Jul/21 $$\mathrm{2}^{\boldsymbol{{a}}!} \:+\:\mathrm{2}^{\boldsymbol{{b}}!} \:+\:\mathrm{2}^{\boldsymbol{{c}}!} \:=\:\boldsymbol{{x}} \\ $$$${Find}\:{natural}\:{numbers}\:\boldsymbol{{a}};\boldsymbol{{b}};\boldsymbol{{c}}\:{such} \\ $$$${that}\:{the}\:{number}\:“\boldsymbol{{x}}''\:{is}\:{a}\:{cube}\:{of} \\ $$$${any}\:{number}. \\ $$ Answered by SinNombre last…

3-x-2-5-x-dx-

Question Number 145259 by mathdanisur last updated on 03/Jul/21 $$\int\:\frac{\left(\mathrm{3}\sqrt{{x}}+\mathrm{2}\right)^{\mathrm{5}} }{\:\sqrt{{x}}}\:{dx}\:=\:? \\ $$ Answered by Dwaipayan Shikari last updated on 03/Jul/21 $${put}\:\sqrt{{x}}={t}\Rightarrow\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}}}{dx}={dt} \\ $$$$=\mathrm{2}.\mathrm{3}^{\mathrm{5}} \int\left({t}+\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{5}}…

Let-a-b-c-0-and-a-2-b-2-c-2-3-Prove-that-1-cyc-a-3-cyc-a-b-3-27-2-a-3-b-3-b-c-3-c-a-3-1-2-c-3-a-b-3-3-For-a-b-c-0-

Question Number 145240 by loveineq last updated on 03/Jul/21 $$\mathrm{Let}\:{a},{b},{c}\:\geqslant\:\mathrm{0}\:\mathrm{and}\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \:=\:\mathrm{3}.\:\mathrm{Prove}\:\mathrm{that} \\ $$$$\left(\mathrm{1}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\underset{{cyc}} {\sum}{a}^{\mathrm{3}} +\underset{{cyc}} {\sum}\left({a}+{b}\right)^{\mathrm{3}} \:\leqslant\:\mathrm{27} \\ $$$$\left(\mathrm{2}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:{a}^{\mathrm{3}} +{b}^{\mathrm{3}} +\left({b}+{c}\right)^{\mathrm{3}} +\left({c}+{a}\right)^{\mathrm{3}} \:\geqslant\:\frac{\mathrm{1}}{\mathrm{2}}\left[{c}^{\mathrm{3}}…

2-3-2-3-2-

Question Number 145224 by mathdanisur last updated on 03/Jul/21 $$\sqrt{\mathrm{2}\sqrt{\mathrm{3}}\:+\:\mathrm{2}}\:-\:\sqrt{\sqrt{\mathrm{3}}\:-\:\sqrt{\mathrm{2}}}\:=\:? \\ $$ Answered by Olaf_Thorendsen last updated on 03/Jul/21 $$\left(\sqrt{\mathrm{3}}+\sqrt{\mathrm{2}}+\mathrm{1}\right)^{\mathrm{2}} \:=\:\mathrm{6}+\mathrm{2}\sqrt{\mathrm{6}}+\mathrm{2}\sqrt{\mathrm{3}}+\mathrm{2}\sqrt{\mathrm{2}} \\ $$$$ \\ $$$${x}\:=\:\sqrt{\mathrm{2}\sqrt{\mathrm{3}}+\mathrm{2}}−\sqrt{\sqrt{\mathrm{3}}−\sqrt{\mathrm{2}}}…

Let-a-b-c-0-c-3-a-b-3-0-and-a-2-b-2-c-2-3-Prove-that-1-2-a-3-b-c-3-b-3-c-a-3-c-3-a-b-3-2-Determine-when-equality-holds-

Question Number 145198 by loveineq last updated on 03/Jul/21 $$\mathrm{Let}\:{a}\geqslant{b}\geqslant{c}\geqslant\mathrm{0}\:,\:{c}^{\mathrm{3}} +\left({a}+{b}\right)^{\mathrm{3}} \neq\mathrm{0}\:\mathrm{and}\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \:=\:\mathrm{3}.\:\mathrm{Prove}\:\mathrm{that} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}\:\leqslant\:\frac{{a}^{\mathrm{3}} +\left({b}+{c}\right)^{\mathrm{3}} +{b}^{\mathrm{3}} +\left({c}+{a}\right)^{\mathrm{3}} }{{c}^{\mathrm{3}} +\left({a}+{b}\right)^{\mathrm{3}} }\:\leqslant\:\mathrm{2} \\ $$$$\mathrm{Determine}\:\mathrm{when}\:\mathrm{equality}\:\mathrm{holds}.…