Find cube root of $63$ to three decimal places

Question

Find $\sqrt[3]{63}$ to three decimal places (without calculus or a calculator).

I am unable to find it. How to get the result by hand calculation? Please help.

Answer 1

$64^{1/3} = 4$, so the answer is a bit less than $4$.
$(4 - x)^3 = 64 - 48 x + 12 x^2 - x^3$. Looking just at the first two terms, we want $x \approx 1/48 \approx 0.021$, i.e. the cube root is approximately $3.979$. This does turn out to be accurate to three decimal places.

Answer 2

$$63^{1/3}=(64-1)^{1/3}=4\left(1-\frac1{64}\right)^{1/3} =4\left(1-\frac{1/3}{64}-\frac{1/9}{64^2}-\cdots\right)$$ (binomial theorem). Surely $4(1-1/(3\times 64))$ will be close enough for your purposes.

Answer 3

Consider $n = a + .1*b + .01*c + .001*d$ where $a,b,c,d$ are integers and $b,c,d$ are all between $-4$ and $5$.

We want to find $a,b,c,d$ so that $n^3$ is close to $63$ as possible.

$n^3 = a^3 + .1(3a^2b) + .01(3a^2c + 3ab^2) + .001(3a^2d + 6abc + b^3) + .0001(6abd + 3ac^2 + 3b^2c) + ....$

$3^3 = 27 < a^3 < 4^3 = 64$. The closest we can get is $a = 4$.

So $63 = a^3 - 1$ and we want to get $.1(3a^2b)$ to be as close to $-1$ as we can. Or $48b \approx -10$. The closest we can get is $b= 0$

So $63 = 64 - 1$ and we want $.01(3a^2c + 3ab^2) = .01(48c)$ to be as close to $-100$ as we can. The closest we can get is $c = -2$ and $.01(48c) = -.96$.

$63 = 64 - .96 -.04$ so we want to get $.001(3a^2d + 6abc + b^3)=.001(48d)$ to be as close $-.04$ as we can. And that would be $d = -1$

I suppose we should check the further more precise value just make sure there is a margin of error that made our choices wrong one.

$64 = 64 - .96 - .048 + .008$ so we want to verify $.0001(6abd + 3ac^2 +3b^2c) = .0001(12) = .00012$ does not alter the approximation by $.01$. It doesn't.

So $n = 4 -.02 - .001 = 3.979$ is the best approximation within $3$ decimal points.

Answer 4

You can try the binary subdivision / Bisection method method which is an equation approximation method which works for any continuous function:

1. We know it will be between 3 and 4, because $3^3 = 27\\4^3=64$ and $f(x)=x^3$ is a monotonically increasing function.
2. Start with step size $0.5$ : $(3.5)^3 = 42.875<63$
3. Halve step size $0.25$ and add it (since we were below): $(3.75)^3 = 52.738<63$
4. Halve step size $0.125$ and add it (since we were below): $(3.875)^3 = 58.186<63$
5. Keep doing this for a while... until the number you last added or subtracted was below $0.001$. We can estimate this with the 2 logarithm that it will happen after $\log_2(10^3) = 9.965 \approx 10$ iterations.