Category: Algebra

Question Number 52406 by Tawa1 last updated on 07/Jan/19 $$\mathrm{Find}\mathrm{the}\mathrm{cube}\mathrm{root}\mathrm{of}26+5\sqrt{3}$$ Answered by MJS last updated on 07/Jan/19 $$\mathrm{no}\mathrm{exact}\mathrm{solution}$$$$\mathrm{I}\mathrm{think}\mathrm{it}\mathrm{should}\mathrm{be}26+15\sqrt{3}$$$$\left({a}+{b}\sqrt{\mathrm{3}}\right)^{\mathrm{3}} =\mathrm{26}+\mathrm{15}\sqrt{\mathrm{3}}…

Question Number 52397 by Joel578 last updated on 07/Jan/19 Commented by Joel578 last updated on 07/Jan/19 Terms of Service Privacy Policy Contact: info@tinkutara.com

Question Number 183381 by Shrinava last updated on 25/Dec/22 $$\frac{{3\mathrm{n}}^5+{4\mathrm{n}}^4-{7\mathrm{n}}^3+{5\mathrm{n}}^2-5}{\mathrm{n}+1}$$$$\mathrm{There}\mathrm{can}\mathrm{be}\mathrm{no}\mathrm{residue}:$$$$\mathrm{a})0\mathrm{b})2\mathrm{c})4\mathrm{d})5\mathrm{e})9$$ Commented by Shrinava last updated…

Question Number 117828 by redouaneee last updated on 13/Oct/20 $$1)(\sqrt{3}-1)(\sqrt{3}+1)=\sqrt{3}\times \sqrt{3}-\sqrt{3}-1$$$$=3-\sqrt{3}-1$$$$=2-\sqrt{3}$$$$2)(2x+\sqrt{3})(2x-\sqrt{3})={\left(2x\right)}^2-2x\sqrt{3}+2x\sqrt{3}-3$$$$=4x^2-3$$ Answered by MJS_new…

Question Number 183363 by Shrinava last updated on 25/Dec/22 $$\mathrm{Find}:$$$$2003\cdot {2005}^3-2004\cdot {2002}^3$$ Answered by Frix last updated on 25/Dec/22 $$\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)\left({x}+\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{3}} −\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)\left({x}−\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{3}} =\mathrm{8}{x}^{\mathrm{3}}…