Cube Roots through Prime Factorization
Interactive Lesson Plan for Class 8 Mathematics
Helps to grab students' attention and activates prior knowledge to set the context.
Quick Quiz: Cube Numbers
What is the cube of 5?
What is the cube of 7?
What is the cube of 10?
Teacher demonstrates the skill or concept clearly, setting a strong example.
Example: Cube root of 3375 using Prime Factorization
Step 1: Find the prime factorization of 3375 using division method:
| 3 | 3375 |
| 3 | 1125 |
| 3 | 375 |
| 5 | 125 |
| 5 | 25 |
| 5 | 5 |
| 1 |
So, 3375 = 3 × 3 × 3 × 5 × 5 × 5 = 3³ × 5³
Therefore, ∛3375 = 3 × 5 = 15
Example 2: Cube root of 729
Prime factorization of 729 using division method:
| 3 | 729 |
| 3 | 243 |
| 3 | 81 |
| 3 | 27 |
| 3 | 9 |
| 3 | 3 |
| 1 |
729 = 3 × 3 × 3 × 3 × 3 × 3 = 3³ × 3³ = (3 × 3)³
So, ∛729 = 3 × 3 = 9
Example 3: Cube root of 216
Prime factorization of 216 using division method:
| 2 | 216 |
| 2 | 108 |
| 2 | 54 |
| 3 | 27 |
| 3 | 9 |
| 3 | 3 |
| 1 |
216 = 2 × 2 × 2 × 3 × 3 × 3 = 2³ × 3³ = (2 × 3)³
So, ∛216 = 2 × 3 = 6
Encourages collaboration, discussion, and peer learning.
Group Activity: Cube Roots from 1 to 20
Work in groups to find the cube roots of these numbers using prime factorization:
1. Find ∛512
Prime factorization using division method:
| 2 | 512 |
| 2 | 256 |
| 2 | 128 |
| 2 | 64 |
| 2 | 32 |
| 2 | 16 |
| 2 | 8 |
| 2 | 4 |
| 2 | 2 |
| 1 |
512 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 2³ × 2³ × 2³ = (2 × 2 × 2)³
So, ∛512 = 2 × 2 × 2 = 8
2. Find ∛1331
Prime factorization using division method:
| 11 | 1331 |
| 11 | 121 |
| 11 | 11 |
| 1 |
1331 = 11 × 11 × 11 = 11³
So, ∛1331 = 11
3. Find ∛2744
Prime factorization using division method:
| 2 | 2744 |
| 2 | 1372 |
| 2 | 686 |
| 7 | 343 |
| 7 | 49 |
| 7 | 7 |
| 1 |
2744 = 2 × 2 × 2 × 7 × 7 × 7 = 2³ × 7³ = (2 × 7)³
So, ∛2744 = 2 × 7 = 14
Allows students to apply learning individually, reinforcing understanding.
Practice Problems
Find the cube root of 8000 using prime factorization.
Prime factorization using division method:
| 2 | 8000 |
| 2 | 4000 |
| 2 | 2000 |
| 2 | 1000 |
| 2 | 500 |
| 2 | 250 |
| 5 | 125 |
| 5 | 25 |
| 5 | 5 |
| 1 |
8000 = 2 × 2 × 2 × 2 × 2 × 2 × 5 × 5 × 5 = 2³ × 2³ × 5³ = (2 × 2 × 5)³
So, ∛8000 = 2 × 2 × 5 = 20
Find the cube root of 13824 using prime factorization.
Prime factorization using division method:
| 2 | 13824 |
| 2 | 6912 |
| 2 | 3456 |
| 2 | 1728 |
| 2 | 864 |
| 2 | 432 |
| 2 | 216 |
| 2 | 108 |
| 2 | 54 |
| 3 | 27 |
| 3 | 9 |
| 3 | 3 |
| 1 |
13824 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 = 2³ × 2³ × 2³ × 3³ = (2 × 2 × 2 × 3)³
So, ∛13824 = 2 × 2 × 2 × 3 = 24
State true or false: For any integer m, m² < m³. Why?
False. This statement is not true for all integers.
For m = 1: 1² = 1 and 1³ = 1 → 1 is not less than 1
For m between 0 and 1: 0.5² = 0.25 and 0.5³ = 0.125 → 0.25 is greater than 0.125
For negative numbers: (-2)² = 4 and (-2)³ = -8 → 4 is greater than -8
The statement is only true for m > 1.
Solve the following problems from the textbook with prime factorization using division method:
1. Find the cube root of each of the following numbers by prime factorisation method.
(i) 64 (ii) 512 (iii) 10648 (iv) 27000 (v) 15625
(vi) 13824 (vii) 110592 (viii) 46656 (ix) 175616 (x) 91125
(i) ∛64
| 2 | 64 |
| 2 | 32 |
| 2 | 16 |
| 2 | 8 |
| 2 | 4 |
| 2 | 2 |
| 1 |
64 = 2 × 2 × 2 × 2 × 2 × 2 = 2³ × 2³ = (2 × 2)³ = 4³
So, ∛64 = 4
(ii) ∛512
| 2 | 512 |
| 2 | 256 |
| 2 | 128 |
| 2 | 64 |
| 2 | 32 |
| 2 | 16 |
| 2 | 8 |
| 2 | 4 |
| 2 | 2 |
| 1 |
512 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 2³ × 2³ × 2³ = (2 × 2 × 2)³ = 8³
So, ∛512 = 8
(iii) ∛10648
| 2 | 10648 |
| 2 | 5324 |
| 2 | 2662 |
| 11 | 1331 |
| 11 | 121 |
| 11 | 11 |
| 1 |
10648 = 2 × 2 × 2 × 11 × 11 × 11 = 2³ × 11³ = (2 × 11)³ = 22³
So, ∛10648 = 22
(iv) ∛27000
| 2 | 27000 |
| 2 | 13500 |
| 2 | 6750 |
| 3 | 3375 |
| 3 | 1125 |
| 3 | 375 |
| 5 | 125 |
| 5 | 25 |
| 5 | 5 |
| 1 |
27000 = 2 × 2 × 2 × 3 × 3 × 3 × 5 × 5 × 5 = 2³ × 3³ × 5³ = (2 × 3 × 5)³ = 30³
So, ∛27000 = 30
(v) ∛15625
| 5 | 15625 |
| 5 | 3125 |
| 5 | 625 |
| 5 | 125 |
| 5 | 25 |
| 5 | 5 |
| 1 |
15625 = 5 × 5 × 5 × 5 × 5 × 5 = 5³ × 5³ = (5 × 5)³ = 25³
So, ∛15625 = 25
Similarly, for the remaining numbers:
(vi) ∛13824 = 24
(vii) ∛110592 = 48
(viii) ∛46656 = 36
(ix) ∛175616 = 56
(x) ∛91125 = 45
2. State true or false.
(i) Cube of any odd number is even.
(ii) A perfect cube does not end with two zeros.
(iii) If square of a number ends with 5, then its cube ends with 25.
(iv) There is no perfect cube which ends with 8.
(v) The cube of a two digit number may be a three digit number.
(vi) The cube of a two digit number may have seven or more digits.
(vii) The cube of a single digit number may be a single digit number.
(i) False - Cube of an odd number is always odd
(ii) True - A perfect cube ends with zeros in multiples of 3
(iii) True - If a number ends with 5, its cube ends with 25
(iv) False - For example, 512 ends with 8 and is a perfect cube
(v) True - For example, 4³ = 64 (two-digit cube of a one-digit number)
(vi) False - The cube of a two-digit number has at most 6 digits
(vii) True - For example, 1³ = 1 and 2³ = 8
Wraps up the lesson with a quick review, reflection, or takeaway to solidify learning.
Quick Self-Assessment
1. What is the cube root of 64?
2. Calculate the cube root of 729 using the prime factorization method.
3. Find the smallest natural number that must be added to 216 to make it a perfect cube.
