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Category: Algebra

To-solve-x-3-x-c-Let-y-x-p-x-3-x-c-dy-dx-x-3-x-c-3x-2-1-x-p-let-dy-dx-mx-3x-2-1-x-p-mx-3x-3-3px-2-m-1-x-p-0-3-x-c-3px-2-m-1-x-p-0-3px-2-m-2-x-3c-p-0-and-since

Question Number 130089 by ajfour last updated on 22/Jan/21 $${To}\:{solve}\:\:\:{x}^{\mathrm{3}} ={x}+{c} \\ $$$${Let}\:\:{y}=\left({x}−{p}\right)\left({x}^{\mathrm{3}} −{x}−{c}\right) \\ $$$$\:\:\frac{{dy}}{{dx}}=\left({x}^{\mathrm{3}} −{x}−{c}\right)+\left(\mathrm{3}{x}^{\mathrm{2}} −\mathrm{1}\right)\left({x}−{p}\right) \\ $$$$\:\:{let}\:\:\:\frac{{dy}}{{dx}}={mx} \\ $$$$\Rightarrow\:\:\left(\mathrm{3}{x}^{\mathrm{2}} −\mathrm{1}\right)\left({x}−{p}\right)={mx} \\ $$$$\mathrm{3}{x}^{\mathrm{3}}…

Question-64542

Question Number 64542 by LPM last updated on 19/Jul/19 Answered by MJS last updated on 19/Jul/19 $${b}=\mathrm{1}−{a} \\ $$$${a}^{\mathrm{2}} +\left(\mathrm{1}−{a}\right)^{\mathrm{2}} =\mathrm{2}\:\Rightarrow\:{a}=\frac{\mathrm{1}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\Rightarrow\:{b}=\frac{\mathrm{1}}{\mathrm{2}}\mp\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$${a}^{\mathrm{7}} =\frac{\mathrm{71}}{\mathrm{16}}\pm\frac{\mathrm{41}\sqrt{\mathrm{3}}}{\mathrm{16}}\:\:\:{b}^{\mathrm{7}} =\frac{\mathrm{71}}{\mathrm{16}}\mp\frac{\mathrm{41}\sqrt{\mathrm{3}}}{\mathrm{16}}…

factorize-x-1-x-3-x-5-x-7-16-

Question Number 64508 by lalitchand last updated on 18/Jul/19 $$\mathrm{factorize}\:\left(\mathrm{x}+\mathrm{1}\right)\left(\mathrm{x}+\mathrm{3}\right)\left(\mathrm{x}+\mathrm{5}\right)\left(\mathrm{x}+\mathrm{7}\right)+\mathrm{16} \\ $$ Commented by Prithwish sen last updated on 18/Jul/19 $$\left(\mathrm{x}+\mathrm{1}\right)\left(\mathrm{x}+\mathrm{7}\right)\left(\mathrm{x}+\mathrm{3}\right)\left(\mathrm{x}+\mathrm{5}\right)+\mathrm{16} \\ $$$$=\left(\mathrm{x}^{\mathrm{2}} +\mathrm{8x}+\mathrm{7}\right)\left(\mathrm{x}^{\mathrm{2}} +\mathrm{8x}+\mathrm{15}\right)+\mathrm{16}…

If-a-b-c-1-a-3-b-3-c-3-4-find-1-a-bc-1-b-ca-1-c-ab-

Question Number 130036 by liberty last updated on 22/Jan/21 $$\mathrm{If}\:\begin{cases}{{a}+{b}+{c}\:=\:\mathrm{1}}\\{{a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} \:=\:\mathrm{4}}\end{cases} \\ $$$$\:\mathrm{find}\:\frac{\mathrm{1}}{{a}+{bc}}\:+\:\frac{\mathrm{1}}{{b}+{ca}}\:+\:\frac{\mathrm{1}}{{c}+{ab}}. \\ $$ Answered by mnjuly1970 last updated on 22/Jan/21 $${solution}:\:\left\{_{{a}^{\mathrm{3}}…

Question-64447

Question Number 64447 by Tawa1 last updated on 18/Jul/19 Answered by som(math1967) last updated on 18/Jul/19 $$\left(\mathrm{4}−\mathrm{3}\right)\left(\mathrm{4}+\mathrm{3}\right)\left(\mathrm{4}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} \right)…..\left(\mathrm{4}^{\mathrm{32}} +\mathrm{3}^{\mathrm{32}} \right)=\mathrm{4}^{{x}} −\mathrm{3}^{{x}} \:\bigstar \\ $$$$\left(\mathrm{4}^{\mathrm{2}}…