Question Number 193026 by Mingma last updated on 02/Jun/23
Answered by aleks041103 last updated on 02/Jun/23
$$\left(\mathrm{3}+\sqrt{\mathrm{5}}\right)^{\mathrm{3}} =\mathrm{3}^{\mathrm{3}} +\mathrm{3}×\mathrm{3}^{\mathrm{2}} ×\sqrt{\mathrm{5}}+\mathrm{3}×\mathrm{3}×\mathrm{5}+\mathrm{5}\sqrt{\mathrm{5}}= \\ $$$$=\mathrm{27}+\mathrm{45}+\mathrm{32}\sqrt{\mathrm{5}}=\mathrm{72}+\mathrm{32}\sqrt{\mathrm{5}} \\ $$$$\mathrm{32}\sqrt{\mathrm{5}}=\sqrt{\mathrm{5}×\mathrm{32}^{\mathrm{2}} }=\sqrt{\mathrm{5}×\mathrm{1024}}=\sqrt{\mathrm{5120}}=\mathrm{71}+{r} \\ $$$$\mathrm{0}<{r}<\mathrm{1} \\ $$$${and}\:{really} \\ $$$$\mathrm{71}^{\mathrm{2}} =\left(\mathrm{70}+\mathrm{1}\right)^{\mathrm{2}} =\mathrm{4900}+\mathrm{1}+\mathrm{140}=\mathrm{5041} \\ $$$$\mathrm{72}^{\mathrm{2}} =\left(\mathrm{70}+\mathrm{2}\right)^{\mathrm{2}} =\mathrm{4900}+\mathrm{4}+\mathrm{280}=\mathrm{5184} \\ $$$$\mathrm{71}^{\mathrm{2}} <\mathrm{5120}<\mathrm{72}^{\mathrm{2}} \Rightarrow\mathrm{71}<\sqrt{\mathrm{5120}}<\mathrm{72} \\ $$$$\Rightarrow\mathrm{143}<\mathrm{72}+\mathrm{32}\sqrt{\mathrm{5}}<\mathrm{144} \\ $$$$\Rightarrow{Ans}.\:\mathrm{143} \\ $$
Commented by Mingma last updated on 02/Jun/23
Perfect ��
Answered by Alleddawi last updated on 02/Jun/23
$$\left(\mathrm{3}+\sqrt{\mathrm{5}}\right)^{\mathrm{3}} =\mathrm{72}+\mathrm{32}\sqrt{\mathrm{5}} \\ $$$$\mathrm{32}\sqrt{\mathrm{5}}=\sqrt{\mathrm{1024}×\mathrm{5}}=\sqrt{\mathrm{5120}} \\ $$$$\mathrm{5041}<\mathrm{5120}<\mathrm{5184} \\ $$$$\sqrt{\mathrm{5041}}<\sqrt{\mathrm{5120}}<\sqrt{\mathrm{5184}} \\ $$$$\mathrm{71}<\sqrt{\mathrm{5120}}<\mathrm{72} \\ $$$$\mathrm{71}<\mathrm{32}\sqrt{\mathrm{5}}<\mathrm{72} \\ $$$$\mathrm{72}+\mathrm{71}<\mathrm{72}+\mathrm{32}\sqrt{\mathrm{5}}<\mathrm{72}+\mathrm{72} \\ $$$$\mathrm{143}<\left(\mathrm{3}+\sqrt{\mathrm{5}}\right)^{\mathrm{3}} <\mathrm{144} \\ $$$$\left(\mathrm{3}+\sqrt{\mathrm{5}}\right)^{\mathrm{3}} \approx\mathrm{143} \\ $$
Commented by Mingma last updated on 03/Jun/23
Perfect ��