Question Number 5771 by Rasheed Soomro last updated on 27/May/16
$$\mathrm{Find}\:\mathrm{to}\:\mathrm{the}\:\mathrm{nearest}\:\mathrm{hundredth}\:\mathrm{the}\:\mathrm{positive} \\ $$$$\mathrm{cube}-\mathrm{root}\:\mathrm{of}\:\mathrm{29}\:. \\ $$
Commented by Yozzii last updated on 27/May/16
$$\left(\mathrm{29}\right)^{\mathrm{1}/\mathrm{3}} =\left(\mathrm{2}+\mathrm{27}\right)^{\mathrm{1}/\mathrm{3}} =\mathrm{3}\left(\mathrm{1}+\frac{\mathrm{2}}{\mathrm{27}}\right)^{\mathrm{1}/\mathrm{3}} \\ $$$${For}\:\mid{x}\mid<\mathrm{1}\:{and}\:{n}\in\mathbb{R},\:{we}\:{can}\:{write} \\ $$$$\left(\mathrm{1}+{x}\right)^{{n}} =\mathrm{1}+{nx}+\frac{{n}\left({n}−\mathrm{1}\right)}{\mathrm{2}!}{x}^{\mathrm{2}} +\frac{{n}\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right)}{\mathrm{3}!}{x}^{\mathrm{3}} +…+\frac{{n}\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right)…\left({n}−{r}+\mathrm{1}\right)}{{r}!}{x}^{{r}} +… \\ $$$$\therefore\:\left(\mathrm{1}+\frac{\mathrm{2}}{\mathrm{27}}\right)^{\mathrm{1}/\mathrm{3}} =\mathrm{1}+\left(\frac{\mathrm{1}}{\mathrm{3}}\right)\left(\frac{\mathrm{2}}{\mathrm{27}}\right)+\frac{\left(\mathrm{1}/\mathrm{3}\right)\left(−\mathrm{2}/\mathrm{3}\right)}{\mathrm{2}!}\left(\frac{\mathrm{2}}{\mathrm{27}}\right)^{\mathrm{2}} +\frac{\left(\mathrm{1}/\mathrm{3}\right)\left(−\mathrm{2}/\mathrm{3}\right)\left(−\mathrm{5}/\mathrm{3}\right)}{\mathrm{3}!}\left(\frac{\mathrm{2}}{\mathrm{27}}\right)^{\mathrm{3}} +… \\ $$$$\left(\mathrm{1}+\frac{\mathrm{2}}{\mathrm{27}}\right)^{\mathrm{1}/\mathrm{3}} \approx\mathrm{1}+\frac{\mathrm{2}}{\mathrm{81}}−\frac{\mathrm{4}}{\mathrm{6561}}+\frac{\mathrm{40}}{\mathrm{1594323}}=\mathrm{1}.\mathrm{0241}\:\left(\mathrm{4}{d}.{p}\right) \\ $$$$\therefore\mathrm{29}^{\mathrm{1}/\mathrm{3}} \approx\mathrm{3}×\mathrm{1}.\mathrm{0241}=\mathrm{3}.\mathrm{07}\:\:\left(\mathrm{2}{d}.{p}\right) \\ $$$$ \\ $$