A cubic number is a number that is obtained when a number is multiplied by taking it three times. Understanding the concept of Cubic numbers is important for students as it will help them in solving various mathematical problems. Cube numbers are also known as perfect cubes. For example, 5 × 5 × 5 = 5³ = 125. A natural number \(n\) is a perfect cube if \(n = {m^3}\) for some natural number \(m\). The cube of an even natural number is an even number, and the cube of an odd natural number is an odd number. Cube of \(1=1×1×1=1\), cube of \(2=2×2×2=8\) and cube of \(3=3×3×3=27\).

Cube Number Definition

The product obtained on multiplying a number three times is called the cube of that number.

Thus, the cube of a natural number is that natural number raised to exponent \(3\).

Example: \(1×1×1=1\). Hence cube of \(1\), i.e. \({1^3} = 1\)

\(2×2×2=8\). Hence cube of \(2\), i.e., \({2^3} = 8\)

For a given number a, cube of a, i.e, \({a^3} = a \times a \times a\)

What is Perfect Cube of Numbers?

The perfect cube of numbers are explained below:

Perfect Cube of Numbers

A number is said to be a perfect cube if it is the cube of some number.

Thus, a natural number \(n\) is a perfect cube if \(n = m \times m \times m = {m^3}\), where \(m\) is a natural number.

Note: Cube of \(0\) is \(0\), and cube of a positive integer is a positive integer.

Cube of a negative integer is a negative integer.

In the table given below, cubes of the first 10 natural numbers have been given.

Number \(a\)	\(1\)	\(2\)	\(3\)	\(4\)	\(5\)	\(6\)	\(7\)	\(8\)	\(9\)	\(10\)
Cube \(\left( {{a^3}} \right)\)	\(1\)	\(8\)	\(27\)	\(64\)	\(125\)	\(216\)	\(343\)	\(512\)	\(729\)	\(1000\)

Cubic Numbers Example

On multiplying a number by itself and then by the number again, the result is a cubic number.The cubes of the first \(20\) natural numbers are given in the following table.

Number \(x\)	Cube \({x^3}\)	Number \(x\)	Cube \({x^3}\)
\(1\)	\(1\)	\(11\)	\(1331\)
\(2\)	\(8\)	\(12\)	\(1728\)
\(3\)	\(27\)	\(13\)	\(2197\)
\(4\)	\(64\)	\(14\)	\(2744\)
\(5\)	\(125\)	\(15\)	\(3375\)
\(6\)	\(216\)	\(16\)	\(4096\)
\(7\)	\(343\)	\(17\)	\(4913\)
\(8\)	\(512\)	\(18\)	\(5832\)
\(9\)	\(729\)	\(19\)	\(6059\)
\(10\)	\(1000\)	\(20\)	\(8000\)

Hardy Ramanujan Number

This story is about one of India’s great mathematicians, S. Ramanujan. Once another famous mathematician Prof. G.H. Hardy came to visit Ramanujan in a taxi whose number was \(1729\). While talking to Ramanujan, Hardy described this number \(1729\)  “a dull number”. Ramanujan quickly pointed out that number \(1729\) was indeed interesting. He said it is the smallest number that can be expressed as a sum of two cubes in two different ways:

\(1729\) is the smallest Hardy Ramanujan Number which can be written as the sum of two cubes in two different ways.

\(1729 = 1728 + 1 = {12^3} + {1^3}\) and \(1729 = 1000 + 729 = {10^3} + {9^3}\)

\(4101, 13832\) are also Hardy Ramanujan Numbers as \(4104 = {2^3} + {16^3}\) or \({9^3} + {15^3}\) and \(13832 = {18^3} + {20^3}\) or \({2^3} + {24^3}.\)

\(1729\) has since been known as the Hardy Ramanujan Number, even though this feature of \(1729\) was known more than 300 years before Ramanujan.

Methods to find Cube of Numbers

Type I: Problems based on Identifying Perfect Cube Numbers

Working Rule:

(i) Write down the given number as the product of prime factors.
(ii) Write the same prime factors together.
(iii) Now, make groups of three equal prime factors.

If no prime factors are left out after grouping as in step (iii), then the number is a perfect cube otherwise, it is not a perfect cube. Thus, if in the prime factorisation of the given number, each factor appears three times, then the number is a perfect cube.

Type II: Problems Based on finding Cubes of Positive Integers

Working Rule:

If \(n\) is a positive integer, then its cube, \(m = n \times n \times n = {n^3}.\)

Type III: Problems based on making a given number a perfect cube:

Working Rule:

1. A given number can be made a perfect cube by multiplying or dividing by a suitable number.
2. Write the given number as the product of prime factors.
3. Make groups of three equal prime factors.
   i) By dividing the given number by the prime factors which are left out, we will get a perfect cube number.
   ii) By multiplying the given number by a suitable number which is such that the prime factors left out in step \(2\) occur three times, we will get a perfect cube number.

Easy Method for finding the Cube of a 2 digit Number

We have: \({(a + b)^3} = {a^3} + 3{a^2}b + 3a{b^2} + {b^3}\)

Method: For finding the cube of a two-digit number with the tens digit \(=a\) and the units digit \(=b\), we make four columns, headed by

\({a^3},\left( {3{a^2} \times b} \right),\left( {3a \times {b^2}} \right)\) and \({b^3}\)

The rest of the procedure is the same as followed in squaring a number by the column method. We simplify the working as

\({a^2} \times a\)	\({a^2} \times 3\,b\)	\({b^2} \times 3\,a\)	\({b^2} \times b\)
\({a^3}\)	\(3\,{a^2}b\)	\(3\,a{b^2}\)	\({b^3}\)

In this method, for each product from the right-hand side, the unit’s digit will be considered, and the rest of the digit(s) will be added to the product preceding it. 

For example: \({27^3} = 19683\) can be calculated as
Answer: 
Here, \(a = 2\) and \(b = 7\)
By using column method, we have

\({2^3} = 8 + 11\)	\(3 \times {2^2} \times 7 = 84 + 32\)	\(3 \times 2 \times {7^2} = 294 + 34\)	\({7^3} = 343\)
\(19\)	\(116\)	\(328\)
\(19\)	\(6\)	\(8\)	\(3\)

Therefore, \({27^3} = 19683.\)

Solved Example Problems on Cubic Numbers

Q.1. Is \(81\) a perfect cube?
Ans: \(81 = 3 \times 3 \times 3 \times 3\)
In the above factorisation \(3\) remains after grouping the \(3’s\) in triplets. Therefore, \(81\) is not a perfect cube.

Q.2. Find \({12^3}\) by using column method.
Ans:

\({a^2} \times a\)	\({a^2} \times 3\,b\)	\({b^2} \times 3\,a\)	\({b^2} \times b\)
\({a^3}\)	\(3\,{a^2}b\)	\(3\,a{b^2}\)	\({b^3}\)

Here, \(a=1\) and \(b=8\).
By using the column method, we have

\({1^3} = 1\)	\(3 \times {1^2} \times 2 = 6 + 1\)	\(3 \times 1 \times {2^2} = 12\)	\({2^3} = 8\)
	\(7\)
\(1\)	\(7\)	\(2\)	\(8\)

Therefore, \({12^3} = 1728\)

Q.3. What is the smallest number by which \(3087\) may be multiplied so that the product is a perfect cube? 
Ans:

Writing \(3087\) as a product of prime factors, we have:
\(3087 = 3 \times 3 \times 7 \times 7 \times 7\)
Clearly, to make it a perfect cube, it must be multiplied by \(3\).

Q.4. Evaluate: \({(2.5)^3}\)
Ans:
\({(2.5)^3} = {\left( {\frac{{25}}{{10}}} \right)^3} = {\left( {\frac{5}{2}} \right)^3} = \frac{{{5^3}}}{{{2^5}}} = \frac{{5 \times 5 \times 5}}{{2 \times 2 \times 2}} = \frac{{125}}{8}\)
Therefore, \({(2.5)^3} = \frac{{125}}{8}\)

Q.5. What is the smallest number by which 1600 must be divided so that the quotient is a perfect cube?
Ans:
The given number is \(1600\).

Writing 1600 as a product of prime factors, we have:
\(1600 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 5 \times 5\)
Clearly, to make it a perfect cube, it must be divided by \(5 \times 5 = 25\).

Summary

In this article, we have learnt about Cube Number Definition, Perfect Cube of Numbers, Cubic Numbers Example, Hardy Ramanujan number, Methods to find cube of numbers. We have also learnt how we can get a perfect cube number by multiplying or dividing it with a suitable number. We have also learnt how we can identify whether any given number is a cube number or not.

Frequently Asked Questions (FAQs) on Cubic Numbers

Q.1. Is \(3\) a cube number?
Ans:
\(3\) is not a cube number because the number \(3\) on prime factorisation gives \(3×1\). Here, the prime factor \(3\) is not in the power of \(3\).

Q.2. What is a cubic number in maths?
Ans: The product obtained on multiplying a number three times is called the cube of that number. 
Thus, the cube of a number is a number raised to power \(3\).
Example: \(1 \times 1 \times 1 = 1\).Hence, cube of \(1\), i.e., \({1^3} = 1\)
\(2 \times 2 \times 2 = 8\). Hence cube of \(2\), i.e., \({2^3} = 8\)
For a given number \(a\), cube of \(a\), i.e., \({a^3} = a \times a \times a\)

Q.3. What are the first \(5\) cube numbers in natural numbers?
Ans:
First \(5\) cube numbers are as follows:
\({1^3} = 1 \times 1 \times 1 = 1\)
\({2^3} = 2 \times 2 \times 2 = 8\)
\({3^3} = 3 \times 3 \times 3 = 27\)
\({4^3} = 4 \times 4 \times 4 = 64\)
\({5^3} = 5 \times 5 \times 5 = 125\)
\({\rm{1,8,27,64}}\) and \(125\) are the first five cube numbers in natural numbers.

Q.4. Which is \({5^{{\rm{th}}}}\) cube number in natural numbers?
Ans:
The cube numbers of natural numbers are
\({1^3} = 1,{2^3} = 8,{3^3} = 27,{4^3} = 64,{5^3} = 125,{6^3} = 216,{7^3} = 343, \ldots \ldots \)
Hence, in natural numbers, \(5th\) cubic number is \(125\).

Q.5. What is Hardy-Ramanujan Number?
Ans: \(1729\) is the smallest Hardy-Ramanujan Number which can be written as the sum of two cubes in two different ways.
\(1729 = 1728 + 1 = {12^3} + {1^3}\) and \(1729 = 1000 + 729 = {10^3} + {9^3}\)

Q.6. What is the cube of three?
Ans: The cube of three is \({3^3} = 3 \times 3 \times 3 = 27\)

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