Category: Algebra

Question Number 203474 by ajfour last updated on 20/Jan/24 $$\frac{{z}^{\mathrm{4}} +\mathrm{4}{z}^{\mathrm{3}} −\mathrm{6}{z}^{\mathrm{2}} −\mathrm{4}{z}+\mathrm{1}}{{z}^{\mathrm{4}} −\mathrm{4}{z}^{\mathrm{3}} −\mathrm{6}{z}^{\mathrm{2}} +\mathrm{4}{z}+\mathrm{1}}=\frac{{z}+\mathrm{1}}{{z}−\mathrm{1}} \\ $$$${Find}\:{z}\in\mathbb{R}. \\ $$ Answered by Rasheed.Sindhi last updated…

Question Number 203497 by ajfour last updated on 20/Jan/24 $${x}^{\mathrm{4}} +{cx}+{d}=\mathrm{0} \\ $$$${then}\:\:{find}\:{p}\:\:{from} \\ $$$${p}^{\mathrm{6}} −\mathrm{4}\left(\frac{{d}^{\:\mathrm{3}} }{{c}^{\mathrm{4}} }\right)^{\mathrm{1}/\mathrm{3}} {p}^{\mathrm{2}} −\mathrm{1}=\mathrm{0} \\ $$$${x}=\frac{{c}^{\mathrm{1}/\mathrm{3}} }{\mathrm{2}}\left({p}\pm\sqrt{−{p}^{\mathrm{2}} −\frac{\mathrm{8}}{{p}}}\:\right) \\…

Question Number 203445 by mathlove last updated on 19/Jan/24 Commented by mr W last updated on 19/Jan/24 $$\Rightarrow{Q}\mathrm{203387} \\ $$ Terms of Service Privacy Policy…

Question Number 203417 by hardmath last updated on 18/Jan/24 $$\mathrm{If} \\ $$$$\mid\overline {−\mathrm{z}\:+\:\boldsymbol{\mathrm{i}}\mathrm{z}}\mid\:=\:\mathrm{2}\:\sqrt{\mathrm{13}}\:\:\:\mathrm{and}\:\:\:\mathrm{z}\:=\:\mathrm{1}\:+\:\left(\mathrm{x}\:+\:\mathrm{1}\right)\centerdot\boldsymbol{\mathrm{i}} \\ $$$$ \\ $$$$\mathrm{Find}:\:\:\:\mathrm{mun}\left(\mathrm{x}\right)\:=\:? \\ $$ Commented by hardmath last updated on…

Question Number 203379 by York12 last updated on 18/Jan/24 $$\mathrm{If}\:{a},{b},{c}\:\in\mathbb{R}^{+} \:\mathrm{with}\:{a}+{b}+{c}=\mathrm{3}\:\mathrm{prove}\:\mathrm{that} \\ $$$$\frac{\mathrm{1}}{{a}^{\mathrm{6}} +{b}^{\mathrm{6}} +\mathrm{3}{c}^{\mathrm{3}} +\mathrm{4}}+\frac{\mathrm{1}}{{b}^{\mathrm{6}} +{c}^{\mathrm{6}} +\mathrm{3}{a}^{\mathrm{3}} +\mathrm{4}}+\frac{\mathrm{1}}{{c}^{\mathrm{6}} +{a}^{\mathrm{6}} +\mathrm{3}{b}^{\mathrm{3}} +\mathrm{4}}\leqslant\frac{\mathrm{3}}{\mathrm{3}+\mathrm{2}\left(\sqrt{{ab}}+\sqrt{{bc}}+\sqrt{{ac}}\right)} \\ $$ Terms…