MathematicsClass 8A Square And A Cube

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.

A SQUARE AND A CUBE

The chapter begins with an intriguing puzzle involving 100 lockers and 100 people, which introduces the concept of factors and their relation to square numbers. Queen Ratnamanjuri's will contains a puzzle where each of the 100 people toggles lockers in a specific pattern: Person 1 toggles every locker, Person 2 toggles every 2nd locker, Person 3 every 3rd locker, and so on until Person 100. The question is which lockers remain open at the end. The key insight is that a locker is toggled once for every factor it has. For example, locker number 6 is toggled by Person 1, 2, 3, and 6, corresponding to its factors. Since factors usually come in pairs (like 1 and 6, 2 and 3), the number of toggles is even, so the locker ends up closed. However, for perfect squares like 1, 4, 9, etc., one factor is repeated (like 2 × 2 for 4), resulting in an odd number of factors and thus toggled an odd number of times, leaving the locker open. Therefore, only lockers numbered with perfect squares remain open. This puzzle elegantly connects the concept of factors, perfect squares, and toggling behavior, setting the stage for the exploration of squares and cubes.

📊 Diagram: The chapter includes diagrams showing the lockers and the toggling process by each person, illustrating how factors correspond to toggling actions. For example, a diagram shows lockers numbered 1 to 100 with some open and some closed after the toggling process.

🧪 Activity: The puzzle itself is an activity where students analyze the toggling pattern and deduce which lockers remain open based on the number of factors.

🔗 Connection: This section leads naturally into the detailed study of square numbers and their properties, as well as the concept of perfect squares having an odd number of factors.

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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