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I have been studying the division method to calculate cube roots by hand (in preparation for an exam in which one is not allowed to use a calculator).

This division method for calculation of cube roots is explained on, e.g., https://www.wikihow.com/Calculate-Cube-Root-by-Hand and https://www.youtube.com/watch?v=y0qWHMmCY4E .

Now, I am trying to calculate $\sqrt[3]{4} $ by hand.

First step, we realize that $1^3 = 1 < 4$ and $2^3 = 8 > 4$, so the leading digit should be a $1$.

Now, it is claimed that the first digit of the solution equals the largest integer, say $x$, such that $300 \cdot 1^2 \cdot x \leq (4-1^3)\cdot 1000 $ which would be 10. Here, I am lost. It should be a 5, but I do not see how to derive that.

I do not understand how to proceed. Can this method be applied in this case? And what would be the reason that this method breaks down for this number?

Any help is much appreciated!

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  • $\begingroup$ This might not be of any help to you on your exam, but outside of that you'd do well to remember that $$\root 3 \of 4 = \root 3 \of {2^2} = (\root 3 \of 2)^2 \approx 1.5874 \approx (1.26)^2.$$ $\endgroup$ – Robert Soupe Oct 16 '18 at 0:13
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Let the approximation of $\sqrt[3]{4000}$ be $10+d$.

Then

$$(10+d)^3=1000+300d+30d^2+d^3\le 4000.$$

We have $15^3=3375$ and $16^3=4096$.

The next digit is given by

$$(150+d)^3=3375000+22500d+450d+d^3\le4000000,$$

$$158^3=3944312,159^3=4019679.$$

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