A Square And A Cube | Class 8 Mathematics Notes
By ConceptScroll Team · Published on 17 July 2026 · 4 min read
A Square And A Cube – this guide gives you a concise, exam-ready overview of A Square And A Cube from Class 8 Mathematics, written by ConceptScroll editors and reviewed against the latest NCERT textbook.
Patterns and Properties of Perfect Squares
This section explores the patterns in the squares of natural numbers, especially focusing on the units digit and differences between consecutive squares. Students are asked to fill in the squares of the first 30 natural numbers and observe patterns such as the units digit of perfect squares always being 0, 1, 4, 5, 6, or 9. The section explains that while the units digit can indicate when a number is not a perfect square (if it ends with 2, 3, 7, or 8), it cannot confirm that a number is a perfect square solely based on its units digit. The section also discusses the pattern of differences between consecutive squares, showing that these differences are consecutive odd numbers (3, 5, 7, 9, etc.). This leads to the important observation that the sum of the first n odd numbers equals n², a fundamental property of perfect squares. Visual proofs are provided to demonstrate this pattern, and students are encouraged to use successive subtraction of odd numbers to determine if a number is a perfect square. The section also prompts students to explore the number of perfect squares within intervals of 100 and their distribution.
📊 Diagram: Diagrams include visual proofs showing how consecutive odd numbers add up to squares using arrangements of unit squares forming larger squares and inverted L shapes.
🧪 Activity: Students are asked to find squares of the first 30 natural numbers, observe patterns, and use successive subtraction of odd numbers to check if numbers are perfect squares.
🔗 Connection: This section leads to the study of square roots and their properties, as well as the introduction of cubes in the next section.
Frequently asked questions
Find the cube roots of these numbers: (i) \(\sqrt[3]{64} =\) (ii) \(\sqrt[3]{512} =\) (iii) \(\sqrt[3]{729} =\)
(i) \(\sqrt[3]{64} = 4\) because \(64 = 4^3\).
(ii) \(\sqrt[3]{512} = 8\) because \(512 = 8^3\).
(iii) \(\sqrt[3]{729} = 9\) because \(729 = 9^3\).
Compute successive differences over levels for perfect cubes until all the differences at a level are the same. What do you notice?
The first level differences of perfect cubes 1, 8, 27, 64, 125, 216 are: 7, 19, 37, 61, 91
The second level differences are: 12, 18, 24, 30
The third level differences are: 6, 6, 6
At the third level, all differences are the same (6).
This shows that the third successive differences of perfect cubes are constant.
1. Find the cube roots of 27000 and 10648. 2. What number will you multiply by 1323 to make it a cube number? 3. State true or false. Explain your reasoning. (i) The cube of any odd number is even. (ii) There is no perfect cube that ends with 8. (iii) The cube of a 2-digit number may be a 3-digit number. (iv) The cube of a 2-digit number may have seven or more digits. (v) Cube numbers have an odd number of factors. 4. You are told that 1331 is a perfect cube. Can you guess without factorisation what its cube root is? Similarly, guess the cube roots of 4913, 12167, and 32768.
1. Cube roots:
- \(\sqrt[3]{27000} = 30\) because \(30^3 = 27000\).
- \(\sqrt[3]{10648} = 22\) because \(22^3 = 10648\).
2. Prime factorise 1323: \(1323 = 3 \times 3 \times 3 \times 7 \times 7 = 3^3 \times 7^2\). To make it a perfect cube, multiply by \(7\) to get \(3^3 \times 7^3\). So, multiply by 7.
3. True or False: (i) False. Cube of odd number is odd. (ii) False. For example, \(2^3 = 8\) ends with 8. (iii) False. Cube of 2-digit number is at least 4-digit number (\(10^3=1000\)). (iv) Tru
Which of the following is the greatest? Explain your reasoning. (i) \(67^{3} - 66^{3}\) (ii) \(43^{3} - 42^{3}\) (iii) \(67^{2} - 66^{2}\) (iv) \(43^{2} - 42^{2}\)
Calculate each difference:
(i) \(67^{3} - 66^{3} = (67 - 66)(67^{2} + 67 \times 66 + 66^{2}) = 1 \times (4489 + 4422 + 4356) = 1 \times 13267 = 13267\).
(ii) \(43^{3} - 42^{3} = (43 - 42)(43^{2} + 43 \times 42 + 42^{2}) = 1 \times (1849 + 1806 + 1764) = 1 \times 5419 = 5419\).
(iii) \(67^{2} - 66^{2} = (67 - 66)(67 + 66) = 1 \times 133 = 133\).
(iv) \(43^{2} - 42^{2} = (43 - 42)(43 + 42) = 1 \times 85 = 85\).
Greatest is (i) \(67^{3} - 66^{3} = 13267\).
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