Question Number 67136 by ,jamiebots last updated on 23/Aug/19 $${factorize}\:\mathrm{2}{x}^{\mathrm{3}} −\mathrm{1} \\ $$ Answered by Cmr 237 last updated on 23/Aug/19 $$\mathrm{2x}^{\mathrm{3}} −\mathrm{1}=\mathrm{2}\left(\mathrm{x}^{\mathrm{3}} −\frac{\mathrm{1}}{\mathrm{2}}\right)=\mathrm{p}\left(\mathrm{x}\right) \\…
Question Number 1585 by Rasheed Soomro last updated on 22/Aug/15 $$\mathrm{Let}\:\omega\:\mathrm{is}\:\mathrm{cube}\:\mathrm{root}\:\mathrm{of}\:\mathrm{unity}\:\mathrm{and}\:\mathrm{x},\mathrm{y},\mathrm{z}\:\in\:\mathbb{Z} \\ $$$$\mathrm{If}\:\omega^{\mathrm{x}} +\omega^{\mathrm{y}} +\omega^{\mathrm{z}} =\mathrm{0}\:\mathrm{prove}\:\mathrm{that}\:\mathrm{3}\:\mid\:\left(\mathrm{x}+\mathrm{y}+\mathrm{z}\right) \\ $$$$\mathrm{Show}\:{by}\:{an}\:{example}\:\mathrm{that}\:\mathrm{the}\:\mathrm{converse}\:\mathrm{is}\:{not}\:\:\mathrm{true}. \\ $$ Commented by 112358 last updated…
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Question Number 1540 by Rasheed Soomro last updated on 17/Aug/15 $$\mathrm{Determine}\:\mathrm{three}\:\mathrm{complex}\:\mathrm{numbers}\:\alpha\:,\:\beta\:,\gamma\:\:\mathrm{such}\:\mathrm{that} \\ $$$$\alpha=\beta^{\:\mathrm{2}} \:\:\:\:\:\:\:{but}\:\:\:\:\beta\:\neq\:\alpha^{\:\mathrm{2}} \\ $$$$\beta\:=\:\gamma^{\:\mathrm{2}} \:\:\:\:\:\:{but}\:\:\:\:\:\gamma\:\neq\:\beta^{\:\mathrm{2}} \\ $$$$\gamma\:=\:\alpha^{\:\mathrm{2}\:} \:\:\:\:\:{but}\:\:\:\:\:\alpha\:\neq\:\gamma^{\:\mathrm{2}} \\ $$ Answered by 123456…
Question Number 1498 by Rasheed Soomro last updated on 14/Aug/15 $$\mathrm{Find}\:\mathrm{complex}\:\mathrm{numbers}\:\alpha\:{and}\:\:\beta\:\:\mathrm{such}\:\mathrm{that} \\ $$$$\alpha^{\:{m}} =\beta^{\:\:{n}} \:\:\:\:{and}\:\:\:\beta^{\:\:{m}} =\alpha^{\mathrm{n}} \:\:,\:{m},{n}\:\in\:\mathbb{Z} \\ $$$$\boldsymbol{\mathrm{D}}\mathrm{etermine}\:\mathrm{formula}\:\mathrm{for}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{pairs}\:\left(\alpha,\beta\right)\:\mathrm{fulfilling}\:\mathrm{the} \\ $$$$\mathrm{above}\:\mathrm{conditions}.\:\:\left(\:\mathrm{You}\:\mathrm{may}\:\mathrm{ignore}\:\mathrm{this}\:\mathrm{part}\:\mathrm{in}\:\mathrm{your}\:\mathrm{answer}\right) \\ $$ Commented by…
Question Number 1497 by Rasheed Soomro last updated on 14/Aug/15 $$\mathrm{Find}\:\mathrm{complex}\:\mathrm{numbers}\:\alpha\:\mathrm{and}\:\beta\:\mathrm{such}\:\mathrm{that} \\ $$$$\alpha^{\mathrm{2}} =\beta^{\mathrm{3}} \:\:\mathrm{and}\:\:\beta^{\mathrm{2}} =\alpha^{\mathrm{3}} \\ $$ Commented by 123456 last updated on 14/Aug/15…
Question Number 67025 by Masumsiddiqui399@gmail.com last updated on 21/Aug/19 Commented by Rasheed.Sindhi last updated on 22/Aug/19 $${x}^{{x}^{{x}+\mathrm{5}} } =\:^{\mathrm{3}} \sqrt{\mathrm{3}} \\ $$$$\left({x}^{{x}^{{x}+\mathrm{5}} } \right)^{\mathrm{3}} =\left(\:^{\mathrm{3}}…
Question Number 1483 by Rasheed Ahmad last updated on 13/Aug/15 $$\mathrm{Prove}\:\mathrm{that}: \\ $$$$\mathrm{a}+\left(\mathrm{b}+\mathrm{c}\right)+\mathrm{d}=\left(\mathrm{a}+\mathrm{c}\right)+\left(\mathrm{d}+\mathrm{b}\right) \\ $$ Commented by prakash jain last updated on 13/Aug/15 $$\mathrm{Basic}\:\mathrm{properties}\:\mathrm{of}\:+\:\mathrm{operator}.\:\mathrm{Or}\:\mathrm{is}\:\mathrm{it} \\…
Question Number 1466 by Rasheed Soomro last updated on 10/Aug/15 $${Without}\:{using}\:{calculus}\:{find}\:{stepwise}\:{solution}: \\ $$$$\underset{{x}\rightarrow−\propto} {{lim}}\:\frac{{a}^{{x}} −\mathrm{1}}{{x}}\:{where}\:{a}>\mathrm{1} \\ $$ Answered by 123456 last updated on 11/Aug/15 $${a}>\mathrm{1},{x}<\mathrm{0}\Rightarrow{a}^{{x}}…
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