Lucas Willems

LUCAS WILLEMS

A 21 year-old student passionate about maths and programming

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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 :

When \(u\) and \(d\) are found, we conclude that \(b = 10d + u\).

The purpose of this article is to show you that Preety Uzlain's technique to calculate cubic root of cube of 2-digits numbers works.

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 !

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