Properties of Cube Numbers
A cube number (or perfect cube) is a number obtained by multiplying a number by itself three times. For example, 8 = 2 × 2 × 2 = 2³ is a perfect cube.
Cube numbers have several interesting properties and patterns related to their units digit, their relationship with odd numbers, and their behaviour under addition and subtraction.
Understanding these properties helps in identifying perfect cubes, finding cube roots mentally, and solving problems in Class 8 NCERT Maths.
While square numbers are symmetrical (the square of a negative number is positive), cube numbers preserve the sign — the cube of a negative number is negative, and the cube of a positive number is positive.
What is Properties of Cube Numbers?
Definition: A number n is called a perfect cube if there exists an integer m such that n = m³ = m × m × m.
Examples of perfect cubes:
- 1³ = 1
- 2³ = 8
- 3³ = 27
- 4³ = 64
- 5³ = 125
- 6³ = 216
- 7³ = 343
- 8³ = 512
- 9³ = 729
- 10³ = 1,000
Notation: The cube of a number m is written as m³.
Properties of Cube Numbers Formula
Cube of a number:
m³ = m × m × m
Sum of first n cubes:
1³ + 2³ + 3³ + ... + n³ = [n(n+1)/2]²
This remarkable formula shows that the sum of the first n cubes equals the square of the sum of the first n natural numbers.
Cube of a sum:
(a + b)³ = a³ + 3a²b + 3ab² + b³
Derivation and Proof
Cube as sum of consecutive odd numbers:
Every perfect cube can be expressed as a sum of consecutive odd numbers:
- 1³ = 1
- 2³ = 3 + 5 = 8
- 3³ = 7 + 9 + 11 = 27
- 4³ = 13 + 15 + 17 + 19 = 64
- 5³ = 21 + 23 + 25 + 27 + 29 = 125
Pattern: The cube of n is the sum of n consecutive odd numbers.
Finding the starting odd number:
- The first odd number in the sum for n³ is: n² − n + 1 = n(n−1) + 1.
- For n = 4: starting odd number = 4 × 3 + 1 = 13. Sum = 13 + 15 + 17 + 19 = 64 = 4³. ✓
Verification of the sum formula:
- 1³ + 2³ + 3³ = 1 + 8 + 27 = 36
- [3 × 4 / 2]² = [6]² = 36 ✓
Types and Properties
The key properties of cube numbers are:
1. Units digit property:
- The units digit of a perfect cube depends on the units digit of the original number.
- If a number ends in 0, its cube ends in 0.
- If it ends in 1, cube ends in 1.
- Ends in 2 → cube ends in 8.
- Ends in 3 → cube ends in 7.
- Ends in 4 → cube ends in 4.
- Ends in 5 → cube ends in 5.
- Ends in 6 → cube ends in 6.
- Ends in 7 → cube ends in 3.
- Ends in 8 → cube ends in 2.
- Ends in 9 → cube ends in 9.
2. Sign property:
- Cube of a positive number is positive.
- Cube of a negative number is negative.
- This is unlike squares, which are always non-negative.
3. Even/Odd property:
- Cube of an even number is even.
- Cube of an odd number is odd.
4. Divisibility property:
- If m is divisible by p, then m³ is divisible by p³.
5. Sum of cubes:
- 1³ + 2³ + ... + n³ = [n(n+1)/2]².
6. Consecutive odd numbers:
- n³ is the sum of n consecutive odd numbers starting from n(n−1) + 1.
- In the prime factorisation of a perfect cube, every prime factor appears a multiple of 3 times.
Solved Examples
Example 1: Example 1: Checking if a number is a perfect cube
Problem: Is 216 a perfect cube?
Solution:
- Prime factorisation of 216:
- 216 = 2 × 108 = 2 × 2 × 54 = 2 × 2 × 2 × 27 = 2³ × 27 = 2³ × 3³
- Each prime factor appears 3 times.
- 216 = (2 × 3)³ = 6³
Answer: Yes, 216 is a perfect cube. 216 = 6³.
Example 2: Example 2: Units digit of a cube
Problem: What is the units digit of 23³?
Solution:
- The units digit of 23 is 3.
- From the property: if a number ends in 3, its cube ends in 7.
- Verification: 23³ = 12,167 (ends in 7) ✓
Answer: The units digit of 23³ is 7.
Example 3: Example 3: Expressing as sum of odd numbers
Problem: Express 5³ as a sum of consecutive odd numbers.
Solution:
- Starting odd number = n(n−1) + 1 = 5 × 4 + 1 = 21
- We need 5 consecutive odd numbers: 21, 23, 25, 27, 29
- Sum = 21 + 23 + 25 + 27 + 29 = 125
- 5³ = 125 ✓
Answer: 5³ = 21 + 23 + 25 + 27 + 29 = 125.
Example 4: Example 4: Sum of first n cubes
Problem: Find 1³ + 2³ + 3³ + 4³ + 5³.
Solution:
- Using the formula: [n(n+1)/2]²
- = [5 × 6 / 2]²
- = [15]²
- = 225
Verification: 1 + 8 + 27 + 64 + 125 = 225 ✓
Answer: The sum is 225.
Example 5: Example 5: Cube of a negative number
Problem: Find (−4)³.
Solution:
- (−4)³ = (−4) × (−4) × (−4)
- = 16 × (−4)
- = −64
Answer: (−4)³ = −64.
Example 6: Example 6: Making a number a perfect cube
Problem: What is the smallest number by which 500 must be multiplied to make it a perfect cube?
Solution:
- Prime factorisation: 500 = 2² × 5³
- For a perfect cube, each prime must appear in groups of 3.
- 2 appears 2 times — needs 1 more 2.
- 5 appears 3 times — already complete.
- Multiply by 2.
- Verification: 500 × 2 = 1,000 = 10³ ✓
Answer: The smallest number is 2.
Example 7: Example 7: Making a number a perfect cube by division
Problem: What is the smallest number by which 1,372 must be divided to make it a perfect cube?
Solution:
- Prime factorisation: 1,372 = 2² × 7³
- 7 appears 3 times — complete.
- 2 appears 2 times — extra 2² = 4.
- Divide by 4.
- Verification: 1,372 / 4 = 343 = 7³ ✓
Answer: The smallest number is 4.
Example 8: Example 8: Predicting units digit
Problem: Without computing, find the units digit of 78³.
Solution:
- The units digit of 78 is 8.
- Property: if a number ends in 8, its cube ends in 2.
Answer: The units digit of 78³ is 2.
Example 9: Example 9: Is 729 a perfect cube?
Problem: Check whether 729 is a perfect cube. If yes, find its cube root.
Solution:
- 729 = 3 × 243 = 3 × 3 × 81 = 3 × 3 × 3 × 27 = 3³ × 3³ = 3⁶
- Each prime factor (3) appears 6 times, which is a multiple of 3.
- 729 = (3²)³ = 9³
Answer: Yes, 729 is a perfect cube. ³√729 = 9.
Example 10: Example 10: Cube of a two-digit number
Problem: Find 15³.
Solution:
- 15³ = 15 × 15 × 15
- 15 × 15 = 225
- 225 × 15 = 225 × 10 + 225 × 5 = 2,250 + 1,125 = 3,375
Verification: Units digit of 15³ should end in 5 (property). 3,375 ends in 5 ✓
Answer: 15³ = 3,375.
Real-World Applications
Finding Cube Roots: Knowing the units digit property helps find cube roots mentally. If a cube ends in 8, the cube root ends in 2.
Prime Factorisation: Checking whether a number is a perfect cube requires verifying that each prime factor appears a multiple of 3 times in its factorisation.
Volume Calculations: The volume of a cube with side a is a³. Understanding cube properties helps in reverse-calculating the side from the volume.
Number Patterns: Cube properties reveal beautiful patterns in mathematics, such as the sum of cubes equalling a perfect square.
Competitive Exams: Knowing cubes of numbers 1–20 and the units digit patterns saves time in aptitude tests and Olympiads.
Key Points to Remember
- A perfect cube n = m³ where m is an integer.
- Cube of a positive number is positive; cube of a negative number is negative.
- Cube of an even number is even; cube of an odd number is odd.
- Units digit pattern: 0→0, 1→1, 2→8, 3→7, 4→4, 5→5, 6→6, 7→3, 8→2, 9→9.
- Every cube can be expressed as a sum of n consecutive odd numbers.
- 1³ + 2³ + ... + n³ = [n(n+1)/2]².
- In the prime factorisation of a perfect cube, every prime appears a multiple of 3 times.
- To make a number a perfect cube, multiply/divide by the smallest factor that completes all prime groups to multiples of 3.
- Cubes of first 10 natural numbers: 1, 8, 27, 64, 125, 216, 343, 512, 729, 1000.
- Hardy-Ramanujan number: 1729 = 12³ + 1³ = 10³ + 9³ (smallest number expressible as sum of two cubes in two ways).
Practice Problems
- Is 5,832 a perfect cube? If yes, find its cube root.
- What is the units digit of 47³?
- Express 6³ as a sum of consecutive odd numbers.
- Find 1³ + 2³ + 3³ + ... + 8³ using the formula.
- What is the smallest number by which 392 must be multiplied to make it a perfect cube?
- What is the smallest number by which 3,456 must be divided to make it a perfect cube?
- Find (−7)³.
- Verify that 1,000 is a perfect cube and find its cube root.
Frequently Asked Questions
Q1. What is a perfect cube?
A perfect cube is a number that can be written as m³, where m is an integer. For example, 27 = 3³ is a perfect cube.
Q2. How do you check if a number is a perfect cube?
Find the prime factorisation. If every prime factor appears a multiple of 3 times, the number is a perfect cube.
Q3. What is the units digit of a cube of a number ending in 7?
If a number ends in 7, its cube ends in 3. For example, 7³ = 343 (ends in 3).
Q4. Can a perfect cube be negative?
Yes. The cube of a negative integer is a negative perfect cube. For example, (−3)³ = −27.
Q5. What is the sum of the first 10 cubes?
1³ + 2³ + ... + 10³ = [10 × 11 / 2]² = 55² = 3,025.
Q6. How is the cube of a number different from the square?
The square of n is n × n (n²); the cube is n × n × n (n³). Squares are always non-negative, but cubes can be negative.
Q7. What is the Hardy-Ramanujan number?
1729. It is the smallest number expressible as the sum of two cubes in two different ways: 1729 = 12³ + 1³ = 10³ + 9³.
Q8. How do you express a cube as a sum of odd numbers?
n³ is the sum of n consecutive odd numbers starting from n(n−1) + 1. For example, 4³ = 13 + 15 + 17 + 19 = 64.
Q9. Are all cube numbers also square numbers?
No. But some numbers are both. For example, 64 = 4³ = 8² and 729 = 9³ = 27². A number that is both a perfect square and a perfect cube is a perfect sixth power.
Q10. Why do we need to learn cube properties?
Cube properties help in mental cube root estimation, checking answers, solving word problems involving volumes, and preparing for higher-class algebra involving cubic polynomials.
