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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.

1.1 Square Numbers

Square numbers are numbers obtained by multiplying a number by itself. The chapter explains why numbers like 1, 4, 9, 16, etc., are called squares by relating them to the area of a square with sides of those lengths. For example, a square with side length 3 units has an area of 3 × 3 = 9 square units. The notation n² is introduced to represent the square of a number n. This notation is read as 'n squared.' The chapter also explains that squares can be formed for fractional and decimal side lengths, such as (3/5)² = 9/25 and (2.5)² = 6.25. The squares of natural numbers are called perfect squares. A table is provided to show the areas of squares with side lengths from 1 to 10 units. The section encourages students to observe patterns in the units digits of perfect squares and to explore the properties of perfect squares, such as the fact that they only end with certain digits (0, 1, 4, 5, 6, or 9) and never with others (2, 3, 7, or 8). This section lays the foundation for understanding the nature of square numbers and their properties.

📊 Diagram: The section includes a table showing side lengths from 1 to 10 units and their corresponding areas (squares). It also shows visual representations of squares with unit squares inside to illustrate the concept of area.

🧪 Activity: Students are encouraged to fill in the squares of the first 30 natural numbers and observe patterns in the units digits and other properties.

🔗 Connection: This section prepares students for the study of patterns and properties of perfect squares, including their relationship with odd numbers and square roots.

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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