Patterns in Cube of Numbers: Exponents are mathematical terms or numbers written as \({a^n}\), where \(a\) is the base and \(n\) denotes the power or exponent to which \(a\) is raised. Exponents raised to power \(2\) are called square numbers or squares, and the exponents raised to power \(2\) are called cubes.

In the cube formula, we multiply a number three times to get its cube. Just as a two-dimensional square can represent square numbers, a cube number can be represented by a three-dimensional cube.

In this article, we will learn about the cubes and the patterns in the cube of a number.

Cube Numbers

Cube numbers can be written as \({a^3} = a \times a \times a\)

If \(a = 1,{a^3} = {1^3} = 1 \times 1 \times 1 = 1\) which can be shown as a cube with length, breadth, and height equal to \(1\) unit.

Fig: Cube Numbers

If \(a = 4,{a^3} = {4^3} = 4 \times 4 \times 4 = 64\) shown by the figure below, a cube with length, breadth, and height equal to \(4\) units.

Have you seen a Rubik’s cube? Look at the picture of Rubik’s cube given below.

It has \(3 \times 3 \times 3\) i.e., \(27\) cubes.

Thus, \(27\) is a cube number.

Thus, the product obtained by multiplying a number three times is called the cube of that number.

? Learn about Cubes and Cube Roots here

How to Find a Perfect Cube?

Now, observe the following table.

Number	Cube	Number	Cube
\(1\)	\(1 \times 1 \times 1 = 1\)	\(6\)	\(6 \times 6 \times 6 = 216\)
\(2\)	\(2 \times 2 \times 2 = 8\)	\(7\)	\(7 \times 7 \times 7 = 343\)
\(3\)	\(3 \times 3 \times 3 = 27\)	\(8\)	\(8 \times 8 \times 8 = 512\)
\(4\)	\(4 \times 4 \times 4 = 64\)	\(9\)	\(9 \times 9 \times 9 = 729\)
\(5\)	\(5 \times 5 \times 5 = 125\)	\(10\)	\(10 \times 10 \times 10 = 1000\)

The cubes obtained in the table are perfect. Thus, a natural number is a perfect cube if it is a cube of some natural number. In other words, a natural number \(n\) is a perfect cube if there exists a natural number \(m\) whose cube is \(n\), i.e., \(n = {m^3}\).

Break down a cube number into its prime factors through the prime factorisation method to identify a perfect cube. We will notice that every factor appears thrice in a number, which is a perfect cube.

Patterns in Cube Numbers

1. Adding Consecutive Odd Numbers

Closely observe the following pattern of sums of odd numbers.

\(1 = 1 = {1^3}\)
\(3 + 5 = 8 = {2^3}\)
\(7 + 9 + 11 = 27 = {3^3}\)
\(13 + 15 + 17 + 19 = 64 = {4^3}\)
\(21 + 23 + 25 + 27 + 29 = 125 = {5^3}\)

2. Cube and Their Prime Factors
Consider the following prime factorisation of the numbers and their cubes.

Prime factorisation of a number	Prime factorisation of its cube
\(4 = 2 \times 2\)	\({4^3} = 64 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 = {2^3} \times {2^3}\)
\(6 = 2 \times 3\)	\({6^3} = 216 = 2 \times 2 \times 2 \times 3 \times 3 \times 3 = {2^3} \times {3^3}\)
\(12 = 2 \times 2 \times 3\)	\({12^3} = 1728 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 = {2^3} \times {2^3} \times {3^3}\)
\(15 = 3 \times 5\)	\({15^3} = 3375 = 3 \times 3 \times 3 \times 5 \times 5 \times 5 = {3^3} \times {5^3}\)

Observe that the prime factor of a number appears three times in the prime factorisation of its cube.

Pattern in Triangular Numbers

Let us look at sums of cubes numbers.

The first cube number is \(1\), which is \({1^3}\) that is also equal to \({1^2}\).

The sum of the first two cube numbers \( = {1^3} + {2^3} = 1 + 8 = 9 = {3^2}\)

The sum of the first three cube numbers \( = {1^3} + {2^3} + {3^3} = 1 + 8 + 27 = 36 = {6^2}\)

The sum of the first four cube numbers \( = {1^3} + {2^3} + {3^3} + {4^3} = 1 + 8 + 27 + 64 = 100 = {10^2}\)

The sum of the first five cube numbers \( = {1^3} + {2^3} + {3^3} + {4^3} + {5^3} = 1 + 8 + 27 + 64 + 125 = 225 = {15^2}\).

Thus, we can say that the sum of the first \(x\) cube numbers is equal to the square of the \({x^{{\rm{th }}}}\) triangular number.

Difference of Cubes of Two Consecutive Positive Integers

Observe the pattern carefully

\({2^3} – {1^3} = 1 + 2 \times 1 \times 3\)
\({3^3} – {2^3} = 1 + 3 \times 2 \times 3\)
\({4^3} – {3^3} = 1 + 4 \times 3 \times 3\)
\({5^3} – {4^3} = 1 + 5 \times 4 \times 3\)

Properties of Cubes

The cube of natural numbers have the following interesting properties:

1. Cubes of all even natural numbers are even.

2. Cubes of all odd natural numbers are odd.

3. The sum of the cubes of first \(n\) natural numbers is equal to the square of their sum.

4. Cubes of the numbers ending with \(4, 5, 6\) and \(9\) are the numbers ending in the same digit. Cubes of numbers ending in digit \(2\) ending in digit \(8\), and the cubes are ending in digit \(8\) ending in digit \(2\). The cubes of the numbers ending in digits \(3\) and \(7\) ends in digits \(7\) and \(3\) respectively.

Computation of Cube Root through a Pattern

In the cube formula, we multiply a number three times to get its cube, so to find the cube root of a number, break down the number to be expressed as a product of three equal numbers, and thus, we get the cube root. Cube root is the number that needs to be multiplied three times to get the original number. Hence, a number \(m\) is the cube root of a number \(n\) if \(n = {m^3}\). In other words, the cube root of a number \(n\) is that number \(m\) whose cube gives \(n\).

We can use the following method to compute the cube root of small numbers, which are perfect cubes of natural numbers.

Let us look at the procedure first.

Step 1: Obtain the natural number.

Step 2: Subtract \(1\) from it. If you get zero as a result, then the cube root of the number is \(1\), else go to the next step.

Step 3: Subtract \(7\left( { = 1 + \frac{{2 \times 1}}{2} \times 6} \right)\) from the resulting number obtained in step 2. If the result is zero, the cube root of the given number is \(2\), else go to the next step.

Step 4: Subtract \(19\left( { = 1 + \frac{{3 \times 2}}{2} \times 6} \right)\) from the resulting number obtained in step 3. If the result is zero, the cube root of the given number is \(3\), else go to the next step.

Step 5: Subtract \(37\left( { = 1 + \frac{{4 \times 3}}{2} \times 6} \right)\) from the resulting number obtained in step 4. If the result is zero, the cube root of the given number is \(4\), else go to the next step.

Step 6: Subtract \(61\left( { = 1 + \frac{{5 \times 4}}{2} \times 6} \right)\) from the resulting number obtained in step 5. If the result is zero, the cube root of the given number is \(5\), else go to the next step.

Continue the process till you get zero. The cube root of the given number will be equal to the number of times the subtraction is performed.

Now, we know that,

\({1^3} = 1 \ldots {\rm{ (i) }}\)
\( \Rightarrow {1^3} – {0^3} = 1 = 1 + 0 \times 6 = 1 + \frac{{1 \times 0}}{2} \times 6\)
\({2^3} – {1^3} = 8 – 1 = 7 \ldots {\rm{ (ii) }}\)
\( \Rightarrow {2^3} – {1^3} = 7 = 1 + 1 \times 6 = 1 + \frac{{2 \times 1}}{2} \times 6\)
\({3^3} – {2^3} = 27 – 8 = 19 \ldots {\rm{ (iii) }}\)
\( \Rightarrow {3^3} – {2^3} = 19 = 1 + 1 \times 6 + 2 \times 6 = 1 + \frac{{3 \times 2}}{2} \times 6\)
\({4^3} – {3^3} = 64 – 27 = 37 \ldots {\rm{ (iv) }}\)
\( \Rightarrow {4^3} – {3^3} = 37 = 1 + 1 \times 6 + 2 \times 6 + 3 \times 6 = 1 + \frac{{4 \times 3}}{2} \times 6\)
\({5^3} – {4^3} = 125 – 64 = 61 \ldots ({\rm{v}})\)
\( \Rightarrow {5^3} – {4^3} = 61 = 1 + 1 \times 6 + 2 \times 6 + 3 \times 6 + 4 \times 6 = 1 + \frac{{5 \times 4}}{2} \times 6\)

Also, \({1^3} = 1\) (By adding equations \((i)\) and \((ii)\))
\({2^3} = 1 + 7\) (By adding equations \((i)\) and \((iii)\)) and so on
The above patterns suggest finding the cube root of a perfect cube of natural number.

Solved Examples – Patterns in Cube of a Numbers

Q.1. Observe the following pattern
\({1^3} = 1\)
\({1^3} + {2^3} = {(1 + 2)^2}\)
\({1^3} + {2^3} + {3^3} = {(1 + 2 + 3)^2}\)
Write the next three rows by using the above pattern.
Ans: The given pattern is
\({1^3} = 1\)
\({1^3} + {2^3} = {(1 + 2)^2}\)
\({1^3} + {2^3} + {3^3} = {(1 + 2 + 3)^2}\)
By observing the above pattern, the next three rows will be;
\({1^3} + {2^3} + {3^3} + {4^3} = {(1 + 2 + 3 + 4)^2}\)
\({1^3} + {2^3} + {3^3} + {4^3} + {5^3} = {(1 + 2 + 3 + 4 + 5)^2}\)
\({1^3} + {2^3} + {3^3} + {4^3} + {5^3} + {6^3} = {(1 + 2 + 3 + 4 + 5 + 6)^2}\)

Q.2. Express \({{\bf{7}}^3}\) as the sum of odd numbers.
Ans: The pattern of sums of odd numbers is as follows.
\(1 = 1 = {1^3}\)
\(3 + 5 = 8 = {2^3}\)
\(7 + 9 + 11 = 27 = {3^3}\)
\(13 + 15 + 17 + 19 = 64 = {4^3}\)
Now, by observing the above pattern, \({{\bf{7}}^3}\) can be written as follows,
\(43 + 45 + 47 + 49 + 51 + 53 + 55 = 343 = {7^3}\)

Q.3. Consider the following pattern.
\({2^3} – {1^3} = 1 + 2 \times 1 \times 3 = 7\)
\({3^3} – {2^3} = 1 + 3 \times 2 \times 3 = 19\)
\({4^3} – {3^3} = 1 + 4 \times 3 \times 3 = 37\)
Now, using the above pattern, find the value of \({51^3} – {50^3}\)
Ans: The given pattern is known as the difference of cubes of two positive integers.
Thus, the value of \({51^3} – {50^3}\) will be,
\({51^3} – {50^3} = 1 + 51 \times 50 \times 3 = 7651\)

Q.4. Is \(216\) a perfect cube? What is that number whose cube is \(216\)?
Ans: Factors of \(216\) are,
\(216 = 2 \times 2 \times 2 \times 3 \times 3 \times 3\)
Now, grouping the factors in triples of equal factors, we get,
\(216 = (2 \times 2 \times 2) \times (3 \times 3 \times 3)\)
We find that the prime factors of \(216\) can be grouped into triples of equal factors, and no factor is left over.
Therefore, \(216\) is a perfect cube.
Thus, taking one factor from each triple, we get,
\(2 \times 3 = 6\)
Hence, \(216\) s the cube of \(6\).

Q.5. Write cubes of the first four positive integers, which can be written in the form \((3\,n + 1)\). For example, \(4, 7, 10\),… and examine the following: “Cube of positive integer which can be written in the form \((3\,n + 1)\) can also be written in this form.”
Ans: Four numbers can be written in form \(3\,n + 1\), when \(n = {\rm{1,2,3,4}}\) are \(4, 7, 10, 13\).
Now, the cube of \(4 = {4^3} = 64 = 3 \times 21 + 1\)
cube of \(7 = {7^3} = 343 = 3 \times 114 + 1\)
cube of \(10 = {10^3} = 1000 = 3 \times 333 + 1\)
cube of \(13 = {13^3} = 2197 = 3 \times 732 + 1\)
Hence, the cube of any positive integer, which can be written as \((3\,n + 1)\)can be written in the form \(3\,n + 1\).

Summary

In this article, we first learnt about the basic concept of finding the cube of a number, and then we learnt about the perfect cube root and the method to find the perfect cube root of a number. In addition to this, we learnt some patterns in the cube of a number followed by the properties of cube numbers. Lastly, we learnt to solve some examples based on the patterns and find the perfect cube to strengthen our hold over the concept.

FAQs

Q.1. Define the perfect cube of a number.
Ans: A natural number is a perfect cube if it is a cube of some natural number. In other words, a natural number \(n\) is a perfect cube if there exists a natural number \(m\) whose cube is \(n\), i.e.,\(n = {m^3}\).

Q.2. What is a number pattern? Explain.
Ans: Numbers are fascinating and astounding because they contain many beautiful patterns and sequences that are incredibly fascinating. A list of numbers with a common trait is called a pattern of numbers. In math, solving problems with number patterns increases a student’s logical thinking and mathematical reasoning ability. To answer any inquiry about whole number patterns in a sequence, the rule used to create the pattern is to be initially understood.

Q.3. What are examples of cube numbers?
Ans: Examples of cube numbers are \(1331,3375,8000,15625\) etc.

Q.4. What are the cube numbers in order?
Ans: The first \(10\) cube number in order are, \({\rm{1,8,27,64,125,216,343,512,729}}\) and \(1000\).

Q.5. Can a perfect cube end with two zeros?
Ans: No, a perfect cube cannot end with two zeros; instead, a perfect cube end with three zeros.

Learn about Volume of Cube here

We hope you find this article on ‘Patterns in Cube of Numbers‘ helpful. In case of any queries, you can reach back to us in the comments section, and we will try to solve them.