# LUCAS WILLEMS

A 25 year-old student passionate about maths and programming

# Calculation of some cubic roots

## Article

I was watching maths videos on Youtube when I came accross this video of Preety Uzlain explaining how to find cube root of cube of 2-digits numbers.

So, if we want to find the value of $$b = \sqrt[3]{a}$$ where $$a$$ is the cube of a 2-digits number, we have to use the following technique :

• Find what is the number $$u \in [[0, 9]]$$ that the cube has the same units digit as $$a$$
• Find what is the number $$d \in [[0, 9]]$$ that the cube is less than $$\left \lfloor \frac{a}{1000} \right \rfloor$$ and the nearest of this value

When $$u$$ and $$d$$ are found, we conclude that $$b = 10d + u$$.

## Proof

As we want to find $$\sqrt[3]{a}$$ where $$a$$ is the cube of 2-digits number, we can write :

$$(d, u) \in [[0, 9]]^2 \qquad a = (10d + u)^3$$

### Proof of the 1st point of the technique

Let's develop the previous expression :

\begin{align}a &= (10d)^3 + 3(10d)^2 u + 3(10d)u^2 + u^3 = 1000d^3 + 300d^2 u + 30du^2 + u^3 \\ &= 10(100d^3 + 30d^2 u + 3du^2) + u^3\end{align}

It is now easy to see that the units digit of $$10(100d^3 + 30d^2 u + 3du^2)$$ is $$0$$, which means that $$a$$ and $$u^3$$ have the same units digit.

So, as the units digit of $$0^3$$, $$1^3$$, $$2^3$$, ..., $$9^3$$ is different, only 1 $$u^3$$ with the same digits unit as $$a$$ exists. It is this reason that proves the 1st point of the technique.

### Proof of the 2nd point of the technique

As $$0 \leqslant u < 10$$, we get the following inequality :

$$10d \leqslant 10d + u < 10d + 10 = 10(d+1)$$And, if we divide by 10 then cube each member of the inequality, we get :

$$d^3 \leqslant \frac{(10d + u)^3}{1000} < (d+1)^3$$

As $$a = (10d + u)^3$$, the final inequality is the following :

$$d^3 \leqslant \frac{a}{1000} < (d+1)^3$$It is now very easy to see that the 2nd point of the technique is right.

Therefore, the technique used by Preety Uzlain works !