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Question Number 5771 by Rasheed Soomro last updated on 27/May/16

$$\mathrm{Find}\mathrm{to}\mathrm{the}\mathrm{nearest}\mathrm{hundredth}\mathrm{the}\mathrm{positive}$$$$\mathrm{cube}-\mathrm{root}\mathrm{of}29.$$

Commented by Yozzii last updated on 27/May/16

$${\left(29\right)}^{1/3}={(2+27)}^{1/3}=3{(1+\frac{2}{27})}^{1/3}$$$$For\mid x\mid <1andn\in \mathbb{R},wecanwrite$$$${(1+x)}^n=1+nx+\frac{n(n-1)}{2!}{x}^2+\frac{n(n-1)(n-2)}{3!}{x}^3+...+\frac{n(n-1)(n-2)...(n-r+1)}{r!}{x}^r+...$$$$\therefore {(1+\frac{2}{27})}^{1/3}=1+\left(\frac{1}{3}\right)\left(\frac{2}{27}\right)+\frac{\left(1/3\right)(-2/3)}{2!}{\left(\frac{2}{27}\right)}^2+\frac{\left(1/3\right)(-2/3)(-5/3)}{3!}{\left(\frac{2}{27}\right)}^3+...$$$${(1+\frac{2}{27})}^{1/3}\approx 1+\frac{2}{81}-\frac{4}{6561}+\frac{40}{1594323}=1.0241(4d.p)$$$$\therefore {29}^{1/3}\approx 3\times 1.0241=3.07(2d.p)$$$$$$

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