Find-to-the-nearest-hundredth-the-positive-cube-root-of-29-

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

Find to the nearest hundredth the positive

cube - root of 29 .

Commented by Yozzii last updated on 27/May/16

{(29)}^{1 / 3} = {(2 + 27)}^{1 / 3} = 3 {(1 + \frac{2}{27})}^{1 / 3}

F o r ∣ x ∣< 1 a n d n \in R, w e c a n w r i t e

{(1 + x)}^{n} = 1 + n x + \frac{n (n - 1)}{2!} x^{2} + \frac{n (n - 1) (n - 2)}{3!} x^{3} + \dots + \frac{n (n - 1) (n - 2) \dots (n - r + 1)}{r!} x^{r} + \dots

∴ {(1 + \frac{2}{27})}^{1 / 3} = 1 + (\frac{1}{3}) (\frac{2}{27}) + \frac{(1 / 3) (- 2 / 3)}{2!} {(\frac{2}{27})}^{2} + \frac{(1 / 3) (- 2 / 3) (- 5 / 3)}{3!} {(\frac{2}{27})}^{3} + \dots

{(1 + \frac{2}{27})}^{1 / 3} \approx 1 + \frac{2}{81} - \frac{4}{6561} + \frac{40}{1594323} = 1.0241 (4 d . p)

∴ 29^{1 / 3} \approx 3 \times 1.0241 = 3.07 (2 d . p)

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